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But let's see what it's saying. It's saying that the composition of f inverse with f has to be equal to the identity matrix. So essentially it's saying, if I apply f to some value in x, if you think about what's this composition doing? This guy's going from x to y. And then this guy goes from y to x. So let's think about what's happening here. f is going from x to y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
This guy's going from x to y. And then this guy goes from y to x. So let's think about what's happening here. f is going from x to y. And then f inverse is going from y to x. So this composition is going to be a mapping from x to x, which the identity matrix, or the identity function, needs to do. It needs to go from x to x. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
f is going from x to y. And then f inverse is going from y to x. So this composition is going to be a mapping from x to x, which the identity matrix, or the identity function, needs to do. It needs to go from x to x. And they're saying this equals identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this right here is that f inverse, the composition of f inverse with f of some member of x, of the set x, is equal to the identity function applied on that item. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
It needs to go from x to x. And they're saying this equals identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this right here is that f inverse, the composition of f inverse with f of some member of x, of the set x, is equal to the identity function applied on that item. These two statements are equivalent. And so by definition, this thing is going to be your original thing. Or another way of writing this is that f inverse applied to f of a is going to be equal to a. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So another way of saying this right here is that f inverse, the composition of f inverse with f of some member of x, of the set x, is equal to the identity function applied on that item. These two statements are equivalent. And so by definition, this thing is going to be your original thing. Or another way of writing this is that f inverse applied to f of a is going to be equal to a. That's what this first statement tells us. And if you think of it visually, it's saying you start with an a, you apply f to it, and you get this value right here, that is f of a. I'm saying it equals b, or I said it equals b earlier on. But then if you apply this f inverse, and it doesn't always exist, but if you apply that f inverse to this function, it needs to go back to this. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Or another way of writing this is that f inverse applied to f of a is going to be equal to a. That's what this first statement tells us. And if you think of it visually, it's saying you start with an a, you apply f to it, and you get this value right here, that is f of a. I'm saying it equals b, or I said it equals b earlier on. But then if you apply this f inverse, and it doesn't always exist, but if you apply that f inverse to this function, it needs to go back to this. By definition, it needs to go back to original a. It has to be equivalent to just doing this little closed loop right here when I introduce you to the identity function. Now that's what this statement is telling us right here. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
But then if you apply this f inverse, and it doesn't always exist, but if you apply that f inverse to this function, it needs to go back to this. By definition, it needs to go back to original a. It has to be equivalent to just doing this little closed loop right here when I introduce you to the identity function. Now that's what this statement is telling us right here. The second statement is saying, look, if I apply f to f inverse, I'm getting the identity function on y. So if I start at some point in y right there, and I apply f inverse first, maybe I go right here. Maybe this point, let's call that lowercase y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Now that's what this statement is telling us right here. The second statement is saying, look, if I apply f to f inverse, I'm getting the identity function on y. So if I start at some point in y right there, and I apply f inverse first, maybe I go right here. Maybe this point, let's call that lowercase y. So this would be f inverse of lowercase y. And then if I were to apply f to that, I know this chart is getting very confusing. If I apply f to this right here, I need to go right back to my original y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Maybe this point, let's call that lowercase y. So this would be f inverse of lowercase y. And then if I were to apply f to that, I know this chart is getting very confusing. If I apply f to this right here, I need to go right back to my original y. So when I apply f to f inverse of y, this has to be equivalent of just doing the identity function on y. So that's what the second statement is saying. Or another way to write it is that f of f inverse of, let's say, of y, where y is a member of the set capital Y, it has to be equal to y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
If I apply f to this right here, I need to go right back to my original y. So when I apply f to f inverse of y, this has to be equivalent of just doing the identity function on y. So that's what the second statement is saying. Or another way to write it is that f of f inverse of, let's say, of y, where y is a member of the set capital Y, it has to be equal to y. And you've been exposed to the idea of an inverse before. We're just doing it a little bit more precisely because we're going to start dealing with these notions, with transformations and matrices in the very near future. So it's good to be exposed to it in this kind of more precise form. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Or another way to write it is that f of f inverse of, let's say, of y, where y is a member of the set capital Y, it has to be equal to y. And you've been exposed to the idea of an inverse before. We're just doing it a little bit more precisely because we're going to start dealing with these notions, with transformations and matrices in the very near future. So it's good to be exposed to it in this kind of more precise form. Now the first thing you might ask is, hey, let's say that I have a function f. Let's say I have a function f. And there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question, is, is f inverse unique? | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So it's good to be exposed to it in this kind of more precise form. Now the first thing you might ask is, hey, let's say that I have a function f. Let's say I have a function f. And there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question, is, is f inverse unique? And actually, probably the obvious question is, how do you know when something's invertible? And we're going to talk a lot about that in the very near future. But let's say we know that f is invertible. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
The obvious question, or maybe it's not an obvious question, is, is f inverse unique? And actually, probably the obvious question is, how do you know when something's invertible? And we're going to talk a lot about that in the very near future. But let's say we know that f is invertible. How do we know, or do we know, whether f inverse is unique? And to answer that question, let's assume it's not unique. So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. So let's say that g is one of them. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
But let's say we know that f is invertible. How do we know, or do we know, whether f inverse is unique? And to answer that question, let's assume it's not unique. So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. So let's say that g is one of them. So let's say g is a mapping. Remember, f is a mapping from x to y. Let's say that g is a mapping from y to x, such that if I apply f to something, and then apply g to it, so this gets me from x to y, and then when I do the composition with g, that gets me back into x. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. So let's say that g is one of them. So let's say g is a mapping. Remember, f is a mapping from x to y. Let's say that g is a mapping from y to x, such that if I apply f to something, and then apply g to it, so this gets me from x to y, and then when I do the composition with g, that gets me back into x. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. So I'm saying that g is an inverse of f. This assumption implies these two things. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Let's say that g is a mapping from y to x, such that if I apply f to something, and then apply g to it, so this gets me from x to y, and then when I do the composition with g, that gets me back into x. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. So I'm saying that g is an inverse of f. This assumption implies these two things. Now let's say that h is another inverse. Let's say that h is another inverse of f. h is another inverse. By definition, by what I just called an inverse, h has to satisfy two requirements. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So I'm saying that g is an inverse of f. This assumption implies these two things. Now let's say that h is another inverse. Let's say that h is another inverse of f. h is another inverse. By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from y to x. And then if I take the composition of h with f, I have to get the identity matrix on the set x. Now that wasn't just part of the definition. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from y to x. And then if I take the composition of h with f, I have to get the identity matrix on the set x. Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The inverse, the composition of the inverse with the function, has to become the identity matrix on x. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The inverse, the composition of the inverse with the function, has to become the identity matrix on x. And then the composition of the function with the inverse has to be the identity function on y. So let's write that. So g is an inverse of f. It implies this. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
The inverse, the composition of the inverse with the function, has to become the identity matrix on x. And then the composition of the function with the inverse has to be the identity function on y. So let's write that. So g is an inverse of f. It implies this. And it also implies that the composition of f with g is equal to the identity matrix. The identity function on y. And if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So g is an inverse of f. It implies this. And it also implies that the composition of f with g is equal to the identity matrix. The identity function on y. And if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well. And just as a reminder, let me redraw what I drew at the beginning, just so we know what we're doing. So if this is the set x right here, let me do it in different color. Let's say this right here is the set y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
And if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well. And just as a reminder, let me redraw what I drew at the beginning, just so we know what we're doing. So if this is the set x right here, let me do it in different color. Let's say this right here is the set y. We know that f is a mapping from x to y. What we're seeing is, what we're trying to determine is, is f's inverse unique? So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Let's say this right here is the set y. We know that f is a mapping from x to y. What we're seeing is, what we're trying to determine is, is f's inverse unique? So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix. So f does that. If you take g, you're going to go back to the same point. So it's equivalent. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix. So f does that. If you take g, you're going to go back to the same point. So it's equivalent. So taking the composition of g with f, that means doing f first, then g. This is equivalent of just taking the identity function in x. So just taking an x and going back to an x. It's equivalent to that. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So it's equivalent. So taking the composition of g with f, that means doing f first, then g. This is equivalent of just taking the identity function in x. So just taking an x and going back to an x. It's equivalent to that. So this is g right here. And the same thing is true with h. h should also be, if I start with some element in x and go into y and then apply h like that, it should also be equivalent to the identity transformation. That's what this statement and this statement are saying. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
It's equivalent to that. So this is g right here. And the same thing is true with h. h should also be, if I start with some element in x and go into y and then apply h like that, it should also be equivalent to the identity transformation. That's what this statement and this statement are saying. Now this statement is saying that if I start with some entry in y here and I apply g, which is the inverse of f, I'm going to go here. So g will take me there. And that when I apply f then to that, I'm going to go back to that same element of y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
That's what this statement and this statement are saying. Now this statement is saying that if I start with some entry in y here and I apply g, which is the inverse of f, I'm going to go here. So g will take me there. And that when I apply f then to that, I'm going to go back to that same element of y. And that's equivalent to just doing the identity function on y. So that's the same thing as the identity function of y. And I could do the same thing here with h. Just take a point here, apply h, then apply f back. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
And that when I apply f then to that, I'm going to go back to that same element of y. And that's equivalent to just doing the identity function on y. So that's the same thing as the identity function of y. And I could do the same thing here with h. Just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
And I could do the same thing here with h. Just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. Can we have two different inverse functions, g and h? So let's start with g. Remember, g is just a mapping from y to x. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from, let me, all these diagrams get confused very quickly. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So let's go back to the question of whether g is unique. Can we have two different inverse functions, g and h? So let's start with g. Remember, g is just a mapping from y to x. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from, let me, all these diagrams get confused very quickly. So let's say this is x and this is y. Remember, g is a mapping from y to x. So g will take us there as a mapping from y to x. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from, let me, all these diagrams get confused very quickly. So let's say this is x and this is y. Remember, g is a mapping from y to x. So g will take us there as a mapping from y to x. And I'm saying that this g is equivalent to the identity mapping or the identity function in composition with this, because all this is saying is you apply g and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So g will take us there as a mapping from y to x. And I'm saying that this g is equivalent to the identity mapping or the identity function in composition with this, because all this is saying is you apply g and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well, by definition, if h is another inverse of f, this is true. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well, by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f and the composition of that with g. And you might want to put parentheses here. Actually, I'll do it very lightly. You might want to put parentheses there. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Well, by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f and the composition of that with g. And you might want to put parentheses here. Actually, I'll do it very lightly. You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions or of transformations is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually, I'll do that. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions or of transformations is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually, I'll do that. I'll put the parentheses there at first, just so you can understand that this is the same thing as that right there, but we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to? The composition of f and g. Well, it's equal to, by definition, it's equal to the identity transformation over y. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Actually, I'll do that. I'll put the parentheses there at first, just so you can understand that this is the same thing as that right there, but we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to? The composition of f and g. Well, it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember, h is a function, is a mapping from y to x. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
The composition of f and g. Well, it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember, h is a function, is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h takes some element in y and gives me some element in x. Now if I take the composition of the identity in y, so that's essentially, I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Remember, h is a function, is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h takes some element in y and gives me some element in x. Now if I take the composition of the identity in y, so that's essentially, I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that. That's the same thing as just applying h to the function to begin with. So just going through this little exercise, we've shown, even though we've started off saying, hey, I have these two different inverses, we've just shown that g must be equal to h. So any function has a unique inverse. You can't set up two different inverses. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
Now if I take the composition of the identity in y, so that's essentially, I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that. That's the same thing as just applying h to the function to begin with. So just going through this little exercise, we've shown, even though we've started off saying, hey, I have these two different inverses, we've just shown that g must be equal to h. So any function has a unique inverse. You can't set up two different inverses. And if you do, you'll find that they're always going to be equal to each other. So so far we know what an inverse is and we don't know what causes someone to be able to have an inverse or not. But we know if they have an inverse, how to think about it, and we also know that that inverse is unique. | Introduction to the inverse of a function Matrix transformations Linear Algebra Khan Academy.mp3 |
What I want to know is, what does this imply about this equation right here? The equation f of x is equal to y. I want to know that for every y that's a member of our codomain, so for every y, let me write this down, for every y that's a member of my codomain, is there a unique, I'll write it in caps, unique solution x that's a member of our domain such that, and I could write such that, well, I'll just write it out. I was going to write it the mathy way, but I think it's nicer to write it in the actual word sometimes. Such that f of x is equal to y. So let me just draw everything out a little bit. We have our set x right here. This is x. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Such that f of x is equal to y. So let me just draw everything out a little bit. We have our set x right here. This is x. We have our codomain y here. We know that f, if you take some point here, let's call that a, it's a member of x. And you apply the function f to it, it'll map you to some element in set y. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
This is x. We have our codomain y here. We know that f, if you take some point here, let's call that a, it's a member of x. And you apply the function f to it, it'll map you to some element in set y. So that's f of a right there. This is so far what this tells us. Now, I want to look at this equation here. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
And you apply the function f to it, it'll map you to some element in set y. So that's f of a right there. This is so far what this tells us. Now, I want to look at this equation here. And I want to know that if I can pick any y in this set, or any lowercase y in this set y, so let's say I pick something here, let's say that's b. I want to know is there a unique solution to the equation f of x is equal to b. Is there a unique solution? So one, I guess you have to think, is there a solution? | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Now, I want to look at this equation here. And I want to know that if I can pick any y in this set, or any lowercase y in this set y, so let's say I pick something here, let's say that's b. I want to know is there a unique solution to the equation f of x is equal to b. Is there a unique solution? So one, I guess you have to think, is there a solution? So is there a solution is saying, look, is there some x here that if I apply the transformation f to it, that I get there? And I also want to know, is it unique? For example, this is the only one that it's unique, but it's not unique if there's some other guy, if there's more than one solution. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So one, I guess you have to think, is there a solution? So is there a solution is saying, look, is there some x here that if I apply the transformation f to it, that I get there? And I also want to know, is it unique? For example, this is the only one that it's unique, but it's not unique if there's some other guy, if there's more than one solution. If there's some other guy in x that if I apply the transformation, I also go to b. This would make it non-unique. Yeah, not unique. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
For example, this is the only one that it's unique, but it's not unique if there's some other guy, if there's more than one solution. If there's some other guy in x that if I apply the transformation, I also go to b. This would make it non-unique. Yeah, not unique. So what I want to concern ourselves with in this video is somehow is invertibility related to the idea of a unique solution to this for any y in our codomain. So let's just work through our definitions of invertibility and see if we can get anywhere constructive. So by definition, f is invertible implies that there exists this little backward looking, three looking thing. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Yeah, not unique. So what I want to concern ourselves with in this video is somehow is invertibility related to the idea of a unique solution to this for any y in our codomain. So let's just work through our definitions of invertibility and see if we can get anywhere constructive. So by definition, f is invertible implies that there exists this little backward looking, three looking thing. This means there exists. I think it's nice to be exposed sometimes to the mathy notations, let me just write there, there exists. That means that there exists some function, let's call it f inverse, that's a mapping from y to x such that, and actually the colons are also the shorthand for such that, but I'll write it out. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So by definition, f is invertible implies that there exists this little backward looking, three looking thing. This means there exists. I think it's nice to be exposed sometimes to the mathy notations, let me just write there, there exists. That means that there exists some function, let's call it f inverse, that's a mapping from y to x such that, and actually the colons are also the shorthand for such that, but I'll write it out. Such that the composition of f inverse with f is equal to the identity on x. So essentially it's saying, look, if I apply f to something in x, and then I apply f inverse to that, I'm going to get back to that point, which is essentially equivalent, or it isn't just essentially equivalent, it is equivalent to just applying the identity function. So that's ix. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
That means that there exists some function, let's call it f inverse, that's a mapping from y to x such that, and actually the colons are also the shorthand for such that, but I'll write it out. Such that the composition of f inverse with f is equal to the identity on x. So essentially it's saying, look, if I apply f to something in x, and then I apply f inverse to that, I'm going to get back to that point, which is essentially equivalent, or it isn't just essentially equivalent, it is equivalent to just applying the identity function. So that's ix. So you just get what you put into it. Such that this inverse function, the composition of the inverse with the function is equal to the identity function, and that the composition of the function with the inverse function is equal to the identity function on y. So if you start in y, and you apply the inverse, and then you apply the function to that, you're going to end up back in y at that same point, and that's equivalent to just applying the identity function. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So that's ix. So you just get what you put into it. Such that this inverse function, the composition of the inverse with the function is equal to the identity function, and that the composition of the function with the inverse function is equal to the identity function on y. So if you start in y, and you apply the inverse, and then you apply the function to that, you're going to end up back in y at that same point, and that's equivalent to just applying the identity function. So this is what invertibility tells me. This is how I defined invertibility in the last video. Now we're concerned with this equation up here. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So if you start in y, and you apply the inverse, and then you apply the function to that, you're going to end up back in y at that same point, and that's equivalent to just applying the identity function. So this is what invertibility tells me. This is how I defined invertibility in the last video. Now we're concerned with this equation up here. We're concerned with the equation, I'll write it in pink, f of x is equal to y. And we want to know for any y, or any lowercase cursive y in our big set y, is there a unique x solution to this? So what we can do is, we know that f is invertible. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Now we're concerned with this equation up here. We're concerned with the equation, I'll write it in pink, f of x is equal to y. And we want to know for any y, or any lowercase cursive y in our big set y, is there a unique x solution to this? So what we can do is, we know that f is invertible. I told you that from the get go. So given that f is invertible, we know that there is this f inverse function. We know that there's this f inverse function, and I can apply that f inverse function. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So what we can do is, we know that f is invertible. I told you that from the get go. So given that f is invertible, we know that there is this f inverse function. We know that there's this f inverse function, and I can apply that f inverse function. It's a mapping from y to x. So I can apply it to any element in y. So for any y, let's say that this is my y right there. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
We know that there's this f inverse function, and I can apply that f inverse function. It's a mapping from y to x. So I can apply it to any element in y. So for any y, let's say that this is my y right there. So I can apply my f inverse to that y, and I'm going to go over here. And of course, y is equal to f of x. These are the exact same points. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So for any y, let's say that this is my y right there. So I can apply my f inverse to that y, and I'm going to go over here. And of course, y is equal to f of x. These are the exact same points. So let's apply our f inverse function to this. So if I apply the f inverse function to both sides of the equation, both sides of this right here is an element in y, and this is the same element in y, right? They're the same element. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
These are the exact same points. So let's apply our f inverse function to this. So if I apply the f inverse function to both sides of the equation, both sides of this right here is an element in y, and this is the same element in y, right? They're the same element. Now, if I apply the mapping, the inverse mapping, to both of that, that's going to take me to some element in x. So let's do that. So if I take the inverse function on both sides of this equation, both sides of this equation, we have some element over here in y, and I'm taking the inverse function to get to some element in x. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
They're the same element. Now, if I apply the mapping, the inverse mapping, to both of that, that's going to take me to some element in x. So let's do that. So if I take the inverse function on both sides of this equation, both sides of this equation, we have some element over here in y, and I'm taking the inverse function to get to some element in x. And what is this going to be equal to? Well, on the right-hand side, we could just write the f inverse of y. That's going to be some element over here. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So if I take the inverse function on both sides of this equation, both sides of this equation, we have some element over here in y, and I'm taking the inverse function to get to some element in x. And what is this going to be equal to? Well, on the right-hand side, we could just write the f inverse of y. That's going to be some element over here. But what does the left-hand side of this equation translate to? The definition of this inverse function is that when you take the composition with f, you're going to end up with the identity function. This is going to be equivalent to, let me write it this way. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
That's going to be some element over here. But what does the left-hand side of this equation translate to? The definition of this inverse function is that when you take the composition with f, you're going to end up with the identity function. This is going to be equivalent to, let me write it this way. This is equal to the composition of f inverse with f of x, which is equivalent to the identity function being applied to x. And then the identity function being applied to x is what? That's just x. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
This is going to be equivalent to, let me write it this way. This is equal to the composition of f inverse with f of x, which is equivalent to the identity function being applied to x. And then the identity function being applied to x is what? That's just x. This thing right here just reduces to x. So we started with the idea that f is invertible. We used the definition of invertibility, that there exists this inverse function right there. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
That's just x. This thing right here just reduces to x. So we started with the idea that f is invertible. We used the definition of invertibility, that there exists this inverse function right there. And then we essentially applied the inverse function to both sides of this equation. Say, look, you give me any y, any lowercase cursive y in this set y, and I will find you a unique x. This is the only x that satisfies this equation. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
We used the definition of invertibility, that there exists this inverse function right there. And then we essentially applied the inverse function to both sides of this equation. Say, look, you give me any y, any lowercase cursive y in this set y, and I will find you a unique x. This is the only x that satisfies this equation. This is the only x. Remember, if, and how do I know it's the only x? Because this is the only possible inverse function. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
This is the only x that satisfies this equation. This is the only x. Remember, if, and how do I know it's the only x? Because this is the only possible inverse function. This is the only possible inverse function. Only one inverse function for which this is true. I proved that to you in the last video. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Because this is the only possible inverse function. This is the only possible inverse function. Only one inverse function for which this is true. I proved that to you in the last video. That if f is invertible, it only has one unique inverse function. We tried before to have maybe two inverse functions, but we saw that they have to be the same thing. So since we only have one inverse function, and it applies to anything in this big uppercase set y, we know we have a solution. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
I proved that to you in the last video. That if f is invertible, it only has one unique inverse function. We tried before to have maybe two inverse functions, but we saw that they have to be the same thing. So since we only have one inverse function, and it applies to anything in this big uppercase set y, we know we have a solution. And because it's only one inverse function, and functions only map to one value in this case, then we know this is a unique solution. So let's write this down. So we've established that if f is invertible, I'll do this in orange, if f is invertible, then the equation f of x is equal to y for all, that little v that looks like it's filled up with something, for all y, the member of our set y has a unique solution. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So since we only have one inverse function, and it applies to anything in this big uppercase set y, we know we have a solution. And because it's only one inverse function, and functions only map to one value in this case, then we know this is a unique solution. So let's write this down. So we've established that if f is invertible, I'll do this in orange, if f is invertible, then the equation f of x is equal to y for all, that little v that looks like it's filled up with something, for all y, the member of our set y has a unique solution. And that unique solution, if you really care about it, is going to be the inverse function applied to y. It might seem like a bit of a no-brainer, but you can see you have to be a little bit precise about it in order to get to the point you want. But let's see if the opposite is true. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So we've established that if f is invertible, I'll do this in orange, if f is invertible, then the equation f of x is equal to y for all, that little v that looks like it's filled up with something, for all y, the member of our set y has a unique solution. And that unique solution, if you really care about it, is going to be the inverse function applied to y. It might seem like a bit of a no-brainer, but you can see you have to be a little bit precise about it in order to get to the point you want. But let's see if the opposite is true. Let's see if we start from the assumption that for all y that is a member of our set y, that the solution, that the equation f of x is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
But let's see if the opposite is true. Let's see if we start from the assumption that for all y that is a member of our set y, that the solution, that the equation f of x is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility. So let's think about it the first way. So we're saying that for any y, so let me draw my sets again, so this is my set x and this is my set y right there. Now we're working for the assumption that you can pick any element in y right here, and then the equation right here has a unique solution. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
If given this, we can prove invertibility. So let's think about it the first way. So we're saying that for any y, so let me draw my sets again, so this is my set x and this is my set y right there. Now we're working for the assumption that you can pick any element in y right here, and then the equation right here has a unique solution. Let's call that unique solution, well, we could call it whatever, but a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that look, you pick a point in y, I can find you some point in x such that f of x is equal to y. And not only can I find that for you, that that is a unique solution. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Now we're working for the assumption that you can pick any element in y right here, and then the equation right here has a unique solution. Let's call that unique solution, well, we could call it whatever, but a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that look, you pick a point in y, I can find you some point in x such that f of x is equal to y. And not only can I find that for you, that that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
And not only can I find that for you, that that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x. And s of, let's say, s of y, where of course y is a member of our set capital Y, s of y is equal to the unique solution in x to f of x is equal to y. Now you're saying, hey Sal, that looks a little convoluted, but think about it, this is a completely valid function definition, right? We're starting with the idea that you give me any y here, you give me any member of this set, and I can always find you a unique solution to this equation. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
It's a mapping from y to x. And s of, let's say, s of y, where of course y is a member of our set capital Y, s of y is equal to the unique solution in x to f of x is equal to y. Now you're saying, hey Sal, that looks a little convoluted, but think about it, this is a completely valid function definition, right? We're starting with the idea that you give me any y here, you give me any member of this set, and I can always find you a unique solution to this equation. Well, OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So why don't I just define a function that says, look, I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
We're starting with the idea that you give me any y here, you give me any member of this set, and I can always find you a unique solution to this equation. Well, OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So why don't I just define a function that says, look, I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here. And of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value, because any value y, any lowercase value y in this set, has a unique solution to f of x is equal to y. So this can only equal one value, so it's well defined. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
That's how I'm defining this function right here. And of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value, because any value y, any lowercase value y in this set, has a unique solution to f of x is equal to y. So this can only equal one value, so it's well defined. So let's apply, let's take some element here. Let's take some element, let me do a good color. Let's say this is b. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So this can only equal one value, so it's well defined. So let's apply, let's take some element here. Let's take some element, let me do a good color. Let's say this is b. And b is a member of y. So let's just map it using our new function right here. So let's take it and map it. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Let's say this is b. And b is a member of y. So let's just map it using our new function right here. So let's take it and map it. And this is s of b right here, which is a member of x. Now, we know that s of b is a unique solution by definition. I know it seems a little circular, but it's not. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So let's take it and map it. And this is s of b right here, which is a member of x. Now, we know that s of b is a unique solution by definition. I know it seems a little circular, but it's not. We know that s of b is a solution. So we know that s of b is a unique solution to f of x is equal to b. Well, if this is the case, we just got this because this is what this function does. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
I know it seems a little circular, but it's not. We know that s of b is a solution. So we know that s of b is a unique solution to f of x is equal to b. Well, if this is the case, we just got this because this is what this function does. It maps every y to the unique solution to this equation. Because we said that every y has a unique solution. So if this is the case, then what happens if I take f of s of b? | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Well, if this is the case, we just got this because this is what this function does. It maps every y to the unique solution to this equation. Because we said that every y has a unique solution. So if this is the case, then what happens if I take f of s of b? Well, I just said this is the unique solution to this. So if I put this guy in here, what am I going to get? I'm going to get b. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So if this is the case, then what happens if I take f of s of b? Well, I just said this is the unique solution to this. So if I put this guy in here, what am I going to get? I'm going to get b. Or another way of saying this is that the composition of f with s applied to b is equal to b. Or another way to say it is that when you take the composition of f with s, this is the same thing. Because if I apply s to b and then I apply f back to that, that's the composition, I just get back to b. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
I'm going to get b. Or another way of saying this is that the composition of f with s applied to b is equal to b. Or another way to say it is that when you take the composition of f with s, this is the same thing. Because if I apply s to b and then I apply f back to that, that's the composition, I just get back to b. That's what's happening here. So this is the same thing as the identity function on y being applied to b. So it's equal to b. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Because if I apply s to b and then I apply f back to that, that's the composition, I just get back to b. That's what's happening here. So this is the same thing as the identity function on y being applied to b. So it's equal to b. So we can say that the composition, we can say that there exists, and we know that this function exists, or that we can always construct this. So we already know that this exists. This existed by me constructing it, but I've hopefully shown you that this is well-defined. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So it's equal to b. So we can say that the composition, we can say that there exists, and we know that this function exists, or that we can always construct this. So we already know that this exists. This existed by me constructing it, but I've hopefully shown you that this is well-defined. From our assumption that this always has a unique solution in x for any y here, I can define this in a fairly reasonable way. So it definitely exists. And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
This existed by me constructing it, but I've hopefully shown you that this is well-defined. From our assumption that this always has a unique solution in x for any y here, I can define this in a fairly reasonable way. So it definitely exists. And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y. Now, let's do another little experiment. Let's take a particular, let me draw our sets again. Let me take some, this is our set x, and let me take some member of set x, call it a. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
And not only does it exist, but we know that the composition of f with this function that I just constructed here is equal to the identity function on y. Now, let's do another little experiment. Let's take a particular, let me draw our sets again. Let me take some, this is our set x, and let me take some member of set x, call it a. Let me take my set y right there. And so we can apply the function to a, and we'll get a member of set y. Let's call that right there, let's call that f of a right there. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Let me take some, this is our set x, and let me take some member of set x, call it a. Let me take my set y right there. And so we can apply the function to a, and we'll get a member of set y. Let's call that right there, let's call that f of a right there. Now, if I apply my magic function here, that always I can get you any member of set y, and I'll give you the unique solution in x to this equation. So let me apply that to this. Let me apply s to this. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Let's call that right there, let's call that f of a right there. Now, if I apply my magic function here, that always I can get you any member of set y, and I'll give you the unique solution in x to this equation. So let me apply that to this. Let me apply s to this. So if I apply s to this, it'll give me the unique solution. Let me write this down. So if I apply s to this, I'm going to apply s to this. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Let me apply s to this. So if I apply s to this, it'll give me the unique solution. Let me write this down. So if I apply s to this, I'm going to apply s to this. And maybe I shouldn't point it back at that. I don't want to imply that it necessarily points back at that. So let me apply s to that. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So if I apply s to this, I'm going to apply s to this. And maybe I shouldn't point it back at that. I don't want to imply that it necessarily points back at that. So let me apply s to that. s to this. So what is this going to point to? What is that point going to be right there? | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So let me apply s to that. s to this. So what is this going to point to? What is that point going to be right there? So that's going to be s of this point, which is f of a, which we know is the unique solution. So this is equal to the unique solution to the equation f of x is equal to this y right here. Or this y right here is just called f of a, right? | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
What is that point going to be right there? So that's going to be s of this point, which is f of a, which we know is the unique solution. So this is equal to the unique solution to the equation f of x is equal to this y right here. Or this y right here is just called f of a, right? Remember, the mapping s just maps you from any member of a to the unique solution to the equation f of x is equal to that, so this is the mapping from f of a to the unique. So this s of f of a is going to be a mapping to the, or this right here is going to be the unique solution to the equation f of x is equal to this member of y. And what's this member of y called? | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Or this y right here is just called f of a, right? Remember, the mapping s just maps you from any member of a to the unique solution to the equation f of x is equal to that, so this is the mapping from f of a to the unique. So this s of f of a is going to be a mapping to the, or this right here is going to be the unique solution to the equation f of x is equal to this member of y. And what's this member of y called? It's called f of a. Well, you could say this in a very convoluted way, but if I were to just, before you learned any linear algebra, if I said, look, if I have the equation f of x is equal to f of a, what is the unique solution to this equation? What does x equal? | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
And what's this member of y called? It's called f of a. Well, you could say this in a very convoluted way, but if I were to just, before you learned any linear algebra, if I said, look, if I have the equation f of x is equal to f of a, what is the unique solution to this equation? What does x equal? Well, x would have to be equal to a. So the unique solution to the equation f of x is equal to f of a is equal to a. And we know that there's only one solution to that. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
What does x equal? Well, x would have to be equal to a. So the unique solution to the equation f of x is equal to f of a is equal to a. And we know that there's only one solution to that. Because that was one of our starting assumptions. So this thing is equal to a. Or we could write s of f of a is equal to a. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
And we know that there's only one solution to that. Because that was one of our starting assumptions. So this thing is equal to a. Or we could write s of f of a is equal to a. Or that the composition of s with f is equal, or applied to a is equal to a, or that the composition of s with f is just the identity function on the set x. This is a mapping right here from x to x. So we could write that the composition of s with f is the identity on x. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Or we could write s of f of a is equal to a. Or that the composition of s with f is equal, or applied to a is equal to a, or that the composition of s with f is just the identity function on the set x. This is a mapping right here from x to x. So we could write that the composition of s with f is the identity on x. So what have we done so far? We started with the idea that you pick any y in our set capital Y here, and we're going to have a unique solution x such that this is true. Such that f of x is equal to y. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
So we could write that the composition of s with f is the identity on x. So what have we done so far? We started with the idea that you pick any y in our set capital Y here, and we're going to have a unique solution x such that this is true. Such that f of x is equal to y. That's what the assumption we started off with. We constructed this function s that immediately maps any member here with its unique solution to this equation. Fair enough. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Such that f of x is equal to y. That's what the assumption we started off with. We constructed this function s that immediately maps any member here with its unique solution to this equation. Fair enough. Now from that, we say, OK, this definitely exists. Not only does it exist, but we figured out that the composition of f with our constructed function is equal to the identity on the set y. And then we also learned that s, the composition of s with f, is the identity function on x. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
Fair enough. Now from that, we say, OK, this definitely exists. Not only does it exist, but we figured out that the composition of f with our constructed function is equal to the identity on the set y. And then we also learned that s, the composition of s with f, is the identity function on x. Let me write this. So we learned this, and we also learned that the composition of f with s is equal to the identity on y. And s clearly exists because I constructed it, and we know it's well-defined because every y, for every y here, there is a solution to this. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |
And then we also learned that s, the composition of s with f, is the identity function on x. Let me write this. So we learned this, and we also learned that the composition of f with s is equal to the identity on y. And s clearly exists because I constructed it, and we know it's well-defined because every y, for every y here, there is a solution to this. So given that I was able to find a function that these two things are true, this is by definition what it means to be invertible. Remember, so this means that f is invertible. Remember, f being invertible, in order for f to be invertible, that means that there must exist some function from, so if f is a mapping from x to y, invertibility means that there must be some function, f inverse, that is a mapping from y to x such that, so I could write there exists a function, such that the inverse function composed with our function should be equal to the identity on x, and the inverse, and the function, and the composition of the function with the inverse function should be the identity on y. | Proof Invertibility implies a unique solution to f(x)=y Linear Algebra Khan Academy.mp3 |