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Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | In the last video, we were able to show that any lambda that satisfies this equation for some non-zero vectors v, then the determinant of lambda times the identity matrix minus a must be equal to 0. Or we could rewrite this as saying lambda is an eigenvalue of a if and only if. I'll write it as if. If and only if the determinant of lambda times the identity matrix minus a is equal to 0. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. So let's do a simple 2 by 2. Let's do an R2. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | If and only if the determinant of lambda times the identity matrix minus a is equal to 0. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. So let's do a simple 2 by 2. Let's do an R2. Let's say that a is equal to the matrix 1, 2, and 4, 3. And I want to find the eigenvalues of a. So if lambda is an eigenvalue of a, let's say lambda is eigenvalue of a, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | Let's do an R2. Let's say that a is equal to the matrix 1, 2, and 4, 3. And I want to find the eigenvalues of a. So if lambda is an eigenvalue of a, let's say lambda is eigenvalue of a, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. So lambda times 1, 0, 0, 1 minus a, 1, 2, 4, 3, is going to be equal to 0. Well, what does this equal to? This right here is the determinant of lambda times this is just lambda times all of these terms. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | So if lambda is an eigenvalue of a, let's say lambda is eigenvalue of a, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. So lambda times 1, 0, 0, 1 minus a, 1, 2, 4, 3, is going to be equal to 0. Well, what does this equal to? This right here is the determinant of lambda times this is just lambda times all of these terms. So it's lambda times 1 is lambda. Lambda times 0 is 0. Lambda times 0 is 0. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | This right here is the determinant of lambda times this is just lambda times all of these terms. So it's lambda times 1 is lambda. Lambda times 0 is 0. Lambda times 0 is 0. Lambda times 1 is lambda. And from that we'll subtract a. So you get 1, 2, 4, 3. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | Lambda times 0 is 0. Lambda times 1 is lambda. And from that we'll subtract a. So you get 1, 2, 4, 3. And this has got to equal 0. And then this matrix, or this difference of matrices, this is just to keep the determinant. This is the determinant of, this first term is going to be lambda minus 1. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | So you get 1, 2, 4, 3. And this has got to equal 0. And then this matrix, or this difference of matrices, this is just to keep the determinant. This is the determinant of, this first term is going to be lambda minus 1. The second term is 0 minus 2. So it's just minus 2. The third term is 0 minus 4. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | This is the determinant of, this first term is going to be lambda minus 1. The second term is 0 minus 2. So it's just minus 2. The third term is 0 minus 4. So it's just minus 4. And then the fourth term is lambda minus 3. Lambda minus 3, just like that. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | The third term is 0 minus 4. So it's just minus 4. And then the fourth term is lambda minus 3. Lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well, everything became a negative. We negated everything. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | Lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well, everything became a negative. We negated everything. And the terms around the diagonal, we had a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | We negated everything. And the terms around the diagonal, we had a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? Well, the determinant of this is just this times that minus this times that. So it's lambda minus 1 times lambda minus 3 minus these two guys multiplied by each other. So minus 2 times minus 4 is plus 8. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | So what's the determinant of this 2 by 2 matrix? Well, the determinant of this is just this times that minus this times that. So it's lambda minus 1 times lambda minus 3 minus these two guys multiplied by each other. So minus 2 times minus 4 is plus 8. Minus 8. This is the determinant of this matrix right here, or this matrix right here, which simplify to that matrix. And that has got to be equal to 0. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | So minus 2 times minus 4 is plus 8. Minus 8. This is the determinant of this matrix right here, or this matrix right here, which simplify to that matrix. And that has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible. And its determinant has to be equal to 0. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | And that has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible. And its determinant has to be equal to 0. So now we have an interesting polynomial equation right here. We can multiply it out. We get what? |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | And its determinant has to be equal to 0. So now we have an interesting polynomial equation right here. We can multiply it out. We get what? Let's multiply it out. We get lambda squared minus 3 lambda plus 3 minus 8 is equal to 0, or lambda squared minus 4 lambda minus 5 is equal to 0. And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | We get what? Let's multiply it out. We get lambda squared minus 3 lambda plus 3 minus 8 is equal to 0, or lambda squared minus 4 lambda minus 5 is equal to 0. And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. Just a little terminology. Polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. Just a little terminology. Polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product as minus 5. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product as minus 5. When you add them, you get minus 4. It's minus 5 and plus 1. So you get lambda minus 5 times lambda plus 1 is equal to 0, right? |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | Let's see, two numbers and you take the product as minus 5. When you add them, you get minus 4. It's minus 5 and plus 1. So you get lambda minus 5 times lambda plus 1 is equal to 0, right? Minus 5 times 1 is minus 5. And then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | So you get lambda minus 5 times lambda plus 1 is equal to 0, right? Minus 5 times 1 is minus 5. And then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. So just like that, using the information that we proved to ourselves in the last video, we're able to figure out that the two eigenvalues of A are lambda equals 5 and lambda equals negative 1. Now that only just solves part of the problem, right? We know we're looking for eigenvalues and eigenvectors. |
Example solving for the eigenvalues of a 2x2 matrix Linear Algebra Khan Academy.mp3 | So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. So just like that, using the information that we proved to ourselves in the last video, we're able to figure out that the two eigenvalues of A are lambda equals 5 and lambda equals negative 1. Now that only just solves part of the problem, right? We know we're looking for eigenvalues and eigenvectors. We know that this equation can be satisfied with lambdas equaling 5 or minus 1. So we know the eigenvalues, but we're yet to determine the actual eigenvectors. So that's what we're going to do in the next video. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So in the last video, I said, look, in standard coordinates, if you have some vector x in your domain and you apply some transformation, then let's say that A is the transformation matrix with respect to the standard basis, then you're just going to have this mapping. You take x, you multiply it by A, you're going to get the transformation of x. Now in the last video and a couple of videos before that, or actually the one right before that, we said, well look, you can do the same mapping but just in an alternate coordinate system. You could do it with respect to some coordinate system or in some coordinate system with respect to some basis B. And that should be the same thing. It should just be a different transformation matrix. And in the last video, we actually figured out what that different transformation matrix is. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | You could do it with respect to some coordinate system or in some coordinate system with respect to some basis B. And that should be the same thing. It should just be a different transformation matrix. And in the last video, we actually figured out what that different transformation matrix is. We had a change of basis. So let's say we had this basis right here. Let me actually copy and paste everything so that we understand what we did. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And in the last video, we actually figured out what that different transformation matrix is. We had a change of basis. So let's say we had this basis right here. Let me actually copy and paste everything so that we understand what we did. So this was the example. Let me copy it, paste it up here, put all of our takeaways from the last video up here. Let me paste it right here. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Let me actually copy and paste everything so that we understand what we did. So this was the example. Let me copy it, paste it up here, put all of our takeaways from the last video up here. Let me paste it right here. So in the last video, we said, OK, this is my basis right there, and then we said, let me copy and paste. Let me copy and paste. That was my alternate basis. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Let me paste it right here. So in the last video, we said, OK, this is my basis right there, and then we said, let me copy and paste. Let me copy and paste. That was my alternate basis. And then I have my change of basis matrix and its inverse. Those will be useful to deal with. So let me copy and paste that. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | That was my alternate basis. And then I have my change of basis matrix and its inverse. Those will be useful to deal with. So let me copy and paste that. OK, copy. And then I'm going to paste it. Edit, paste. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So let me copy and paste that. OK, copy. And then I'm going to paste it. Edit, paste. Maybe I'll just write it over there. Not maybe the best order. Maybe I should have written that first, but I think we get the idea. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Edit, paste. Maybe I'll just write it over there. Not maybe the best order. Maybe I should have written that first, but I think we get the idea. And then we want to figure out, we want to write what our transformation matrix is with respect to the standard basis. And I wrote that right over here. This was all from the last problem, if you're wondering where I got all this stuff. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Maybe I should have written that first, but I think we get the idea. And then we want to figure out, we want to write what our transformation matrix is with respect to the standard basis. And I wrote that right over here. This was all from the last problem, if you're wondering where I got all this stuff. Let me copy and then paste that. Edit, paste. So I'll paste that. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | This was all from the last problem, if you're wondering where I got all this stuff. Let me copy and then paste that. Edit, paste. So I'll paste that. And then the whole point of the last video is we figured out what the transformation matrix is with respect to this basis, right here. So D, which was the big takeaway from the last video, was equal to this right here. Let me copy and paste that. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So I'll paste that. And then the whole point of the last video is we figured out what the transformation matrix is with respect to this basis, right here. So D, which was the big takeaway from the last video, was equal to this right here. Let me copy and paste that. And now we have all of our takeaways in one place. Edit, paste. What I want to do in this video is to verify that D actually works. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Let me copy and paste that. And now we have all of our takeaways in one place. Edit, paste. What I want to do in this video is to verify that D actually works. That I could start with some vector x that I can, let me write it up here. Let's take some example vector. So this transformation, its entire domain is R2. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | What I want to do in this video is to verify that D actually works. That I could start with some vector x that I can, let me write it up here. Let's take some example vector. So this transformation, its entire domain is R2. So let's start with some vector x. Let's say that x is equal to 1 minus 1. Now, we could just apply the transformation in the traditional way and get the transformation of x. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So this transformation, its entire domain is R2. So let's start with some vector x. Let's say that x is equal to 1 minus 1. Now, we could just apply the transformation in the traditional way and get the transformation of x. So let's just do that. The transformation of x is just this matrix times x. And so what is that going to equal? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Now, we could just apply the transformation in the traditional way and get the transformation of x. So let's just do that. The transformation of x is just this matrix times x. And so what is that going to equal? Let me see, maybe I can just do it right here in this corner to save space. So it's going to be this matrix times x. So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And so what is that going to equal? Let me see, maybe I can just do it right here in this corner to save space. So it's going to be this matrix times x. So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right? Minus 2 times minus 1 is just 2. So it's going to be 3 plus 2. So it's going to be equal to 5. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right? Minus 2 times minus 1 is just 2. So it's going to be 3 plus 2. So it's going to be equal to 5. And then the second term right here is going to be 2 times 1 plus minus 2 times minus 1. Well, that's just positive 2. So it's going to be 2 plus 2. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So it's going to be equal to 5. And then the second term right here is going to be 2 times 1 plus minus 2 times minus 1. Well, that's just positive 2. So it's going to be 2 plus 2. So that is 4. So that's just the transformation of x. Now, what is this vector x represented in coordinates, or I guess you could say to this alternate basis coordinates? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So it's going to be 2 plus 2. So that is 4. So that's just the transformation of x. Now, what is this vector x represented in coordinates, or I guess you could say to this alternate basis coordinates? So what is that vector x represented in coordinates with respect to this basis right here? Well, you saw before, I wrote them out here. Maybe it'll be useful to do it right here. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Now, what is this vector x represented in coordinates, or I guess you could say to this alternate basis coordinates? So what is that vector x represented in coordinates with respect to this basis right here? Well, you saw before, I wrote them out here. Maybe it'll be useful to do it right here. The coordinate, I'll copy this. Actually, let me copy both of these. These will both be useful. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Maybe it'll be useful to do it right here. The coordinate, I'll copy this. Actually, let me copy both of these. These will both be useful. Edit, copy. As you can see, if you want to go from x to the x in an alternate basis, or the alternate coordinate representations of x, you essentially multiply x times c inverse. But that's why I'm copying and pasting it. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | These will both be useful. Edit, copy. As you can see, if you want to go from x to the x in an alternate basis, or the alternate coordinate representations of x, you essentially multiply x times c inverse. But that's why I'm copying and pasting it. Let me copy and then let me put it up here so that we can apply these. So then paste it right there. So if we want to go from x to the b coordinates of x, I take my x and I multiply it times c inverse. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | But that's why I'm copying and pasting it. Let me copy and then let me put it up here so that we can apply these. So then paste it right there. So if we want to go from x to the b coordinates of x, I take my x and I multiply it times c inverse. c inverse is this thing right here. So if I take x and I multiply it times c inverse, I'll get this version of x. So let's do that. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So if we want to go from x to the b coordinates of x, I take my x and I multiply it times c inverse. c inverse is this thing right here. So if I take x and I multiply it times c inverse, I'll get this version of x. So let's do that. So this times that, let me just put the minus 1 third out front, so it's going to be equal to minus 1 third times, let's see if we can do this one in our head as well. So it's going to be 1 times 1 plus minus 2 times minus 1, which is just positive 2. So it's going to be 1 plus 2. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So let's do that. So this times that, let me just put the minus 1 third out front, so it's going to be equal to minus 1 third times, let's see if we can do this one in our head as well. So it's going to be 1 times 1 plus minus 2 times minus 1, which is just positive 2. So it's going to be 1 plus 2. So it's going to be equal to 3. And then it's minus 2 times 1, which is minus 2, plus 1 times minus 1, which is just minus 1. So it's minus 2 minus 1. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So it's going to be 1 plus 2. So it's going to be equal to 3. And then it's minus 2 times 1, which is minus 2, plus 1 times minus 1, which is just minus 1. So it's minus 2 minus 1. It's minus 3. So if we have minus 1 third times this, the b coordinate representation of our vector x is going to be equal to minus 1 and then 1, just like that. Which is actually interesting for this example. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So it's minus 2 minus 1. It's minus 3. So if we have minus 1 third times this, the b coordinate representation of our vector x is going to be equal to minus 1 and then 1, just like that. Which is actually interesting for this example. It just kind of swapped the first entry and the second entry. Now let's see what happens when we apply d to x. So if we apply d to x, d should be our transformation matrix if we're dealing in the b coordinates. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Which is actually interesting for this example. It just kind of swapped the first entry and the second entry. Now let's see what happens when we apply d to x. So if we apply d to x, d should be our transformation matrix if we're dealing in the b coordinates. So let's see what happens. So if we apply d to x, let me scroll over a little bit, just so that we get a little bit more real estate. So if we apply d to x, what do we get? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So if we apply d to x, d should be our transformation matrix if we're dealing in the b coordinates. So let's see what happens. So if we apply d to x, let me scroll over a little bit, just so that we get a little bit more real estate. So if we apply d to x, what do we get? And so this is going to be the transformation, or this should be the transformation of x in b coordinates. So what is this going to be equal to? We have to multiply this times d. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So if we apply d to x, what do we get? And so this is going to be the transformation, or this should be the transformation of x in b coordinates. So what is this going to be equal to? We have to multiply this times d. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1. So it's just minus 1 times minus 1, which is 1. And then we're going to get 0 times minus 1 plus 2 times 1. So 2 times 1 is just 2. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | We have to multiply this times d. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1. So it's just minus 1 times minus 1, which is 1. And then we're going to get 0 times minus 1 plus 2 times 1. So 2 times 1 is just 2. Now, for everything to work together, and assuming I haven't made any careless mistakes, this thing, this vector right here, should be the same as this vector if I change my basis. So if I go from the standard basis to the basis b, and when you go in that direction, you just multiply this guy times c inverse. So I'm just using this formula right here. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So 2 times 1 is just 2. Now, for everything to work together, and assuming I haven't made any careless mistakes, this thing, this vector right here, should be the same as this vector if I change my basis. So if I go from the standard basis to the basis b, and when you go in that direction, you just multiply this guy times c inverse. So I'm just using this formula right here. If I'm in the standard basis, I multiply by c inverse, I'm going to get the b basis. So let's see what I get. So this guy, I'm going to multiply him times c inverse. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So I'm just using this formula right here. If I'm in the standard basis, I multiply by c inverse, I'm going to get the b basis. So let's see what I get. So this guy, I'm going to multiply him times c inverse. Let me do it up here just to get some extra space. Do it right here. So I'm going to multiply the vector 5, 4. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So this guy, I'm going to multiply him times c inverse. Let me do it up here just to get some extra space. Do it right here. So I'm going to multiply the vector 5, 4. I'm going to multiply that by c inverse. We're going to have minus 1 third times 1 minus 2 minus 2 1, just like that. So this is going to be equal to, I'll just write the minus 1 third out front. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So I'm going to multiply the vector 5, 4. I'm going to multiply that by c inverse. We're going to have minus 1 third times 1 minus 2 minus 2 1, just like that. So this is going to be equal to, I'll just write the minus 1 third out front. And we have 1 times 5, which is 5. Let me just write this right. 5 plus minus 2 times 4. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | So this is going to be equal to, I'll just write the minus 1 third out front. And we have 1 times 5, which is 5. Let me just write this right. 5 plus minus 2 times 4. So 5 minus 8. And then we have minus 2 times 5, which is minus 10. And then we have plus 1 times 4. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | 5 plus minus 2 times 4. So 5 minus 8. And then we have minus 2 times 5, which is minus 10. And then we have plus 1 times 4. Minus 2 times 5, which is minus 10 plus 1 times 4. So this is equal to minus 1 third times minus 3. And this is what? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And then we have plus 1 times 4. Minus 2 times 5, which is minus 10 plus 1 times 4. So this is equal to minus 1 third times minus 3. And this is what? This is minus 6. If you multiply the minus 1 third times, all the negatives cancel out and you get 1 and 2. Which is exactly what we needed to get. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And this is what? This is minus 6. If you multiply the minus 1 third times, all the negatives cancel out and you get 1 and 2. Which is exactly what we needed to get. When you take this guy and you change its basis to basis b, or you change its coordinate system to the coordinate system with respect to b, you multiply it by c inverse, you get that right there. So this literally is the b coordinate representation of the transformation of x. We just did it by multiplying it by c inverse. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Which is exactly what we needed to get. When you take this guy and you change its basis to basis b, or you change its coordinate system to the coordinate system with respect to b, you multiply it by c inverse, you get that right there. So this literally is the b coordinate representation of the transformation of x. We just did it by multiplying it by c inverse. Which is exactly what we got when we took the b coordinate version of x and we applied that matrix that we found, that transformation matrix with respect to the b coordinates, and you multiply it times this guy right here, we got the same answer. So it didn't matter whether we went this way around this little cycle or we went this way, we got the same answer. This isn't a proof, but it shows us that what we did in the last video at least works for this case. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | We just did it by multiplying it by c inverse. Which is exactly what we got when we took the b coordinate version of x and we applied that matrix that we found, that transformation matrix with respect to the b coordinates, and you multiply it times this guy right here, we got the same answer. So it didn't matter whether we went this way around this little cycle or we went this way, we got the same answer. This isn't a proof, but it shows us that what we did in the last video at least works for this case. And I literally did pick a random x here, and you can verify it if you like for other things. Now, you should hopefully be reasonably convinced that we can do this, that you can change your basis and find a transformation matrix. We've shown how to do it. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | This isn't a proof, but it shows us that what we did in the last video at least works for this case. And I literally did pick a random x here, and you can verify it if you like for other things. Now, you should hopefully be reasonably convinced that we can do this, that you can change your basis and find a transformation matrix. We've shown how to do it. But the obvious question is, why do you do it? Why? Why do you do it? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | We've shown how to do it. But the obvious question is, why do you do it? Why? Why do you do it? And someone actually wrote a comment on the last video, which I think is actually, it kind of captures the art of why you do it. I'm not looking at the comment right now, but if I remember correctly, they said their linear algebra teacher said that linear algebra is the art of choosing the right basis. Let me write that down. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Why do you do it? And someone actually wrote a comment on the last video, which I think is actually, it kind of captures the art of why you do it. I'm not looking at the comment right now, but if I remember correctly, they said their linear algebra teacher said that linear algebra is the art of choosing the right basis. Let me write that down. The art of choosing the right basis. Or you could imagine the right coordinate system. And why is there a right coordinate system? |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Let me write that down. The art of choosing the right basis. Or you could imagine the right coordinate system. And why is there a right coordinate system? Maybe I'll put little quotes inside the quotation. What does it mean to have the right coordinate system? Well, if you look at the original transformation matrix with respect to the standard basis, it's fine. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And why is there a right coordinate system? Maybe I'll put little quotes inside the quotation. What does it mean to have the right coordinate system? Well, if you look at the original transformation matrix with respect to the standard basis, it's fine. It's got this 2 by 2. But if you performed matrix operations with this, you've got to do some math. And if you had to perform it over and over, if you had to perform it on a bunch of vectors, it might get a little bit, you know, it is what it is. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Well, if you look at the original transformation matrix with respect to the standard basis, it's fine. It's got this 2 by 2. But if you performed matrix operations with this, you've got to do some math. And if you had to perform it over and over, if you had to perform it on a bunch of vectors, it might get a little bit, you know, it is what it is. But when you transfer your bases, when you go to a new basis, when you went to this basis right here, all of a sudden you find that the transformation matrix is much simpler. It's a diagonal matrix. When you multiply a diagonal matrix times something, you're literally just taking scaling factors of the first and second terms. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And if you had to perform it over and over, if you had to perform it on a bunch of vectors, it might get a little bit, you know, it is what it is. But when you transfer your bases, when you go to a new basis, when you went to this basis right here, all of a sudden you find that the transformation matrix is much simpler. It's a diagonal matrix. When you multiply a diagonal matrix times something, you're literally just taking scaling factors of the first and second terms. When you multiply this guy times some vector, we did it here, when you multiplied this guy times this vector, you literally just scaled the first term times minus 1, and you scaled the second term by 2. So it's a much simpler operation. And you might say, hey, but we have to do all that work of multiplying by c inverse to get there, and then once you get this answer, you're going to have to multiply by c to go back into standard coordinates. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | When you multiply a diagonal matrix times something, you're literally just taking scaling factors of the first and second terms. When you multiply this guy times some vector, we did it here, when you multiplied this guy times this vector, you literally just scaled the first term times minus 1, and you scaled the second term by 2. So it's a much simpler operation. And you might say, hey, but we have to do all that work of multiplying by c inverse to get there, and then once you get this answer, you're going to have to multiply by c to go back into standard coordinates. You know, that's a lot more work than just what you save here. But imagine if you had to apply d multiple times. Imagine if you had to apply d times d times d times d to x. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | And you might say, hey, but we have to do all that work of multiplying by c inverse to get there, and then once you get this answer, you're going to have to multiply by c to go back into standard coordinates. You know, that's a lot more work than just what you save here. But imagine if you had to apply d multiple times. Imagine if you had to apply d times d times d times d to x. Or, you know, let me say it this way. Imagine if you had to apply a times a times a. Like, you have to apply a 100 times to some vector up here. |
Alternate basis transformation matrix example part 2 Linear Algebra Khan Academy.mp3 | Imagine if you had to apply d times d times d times d to x. Or, you know, let me say it this way. Imagine if you had to apply a times a times a. Like, you have to apply a 100 times to some vector up here. Then applying a 100 times to some vector, it would be much more computationally intensive than applying d 100 times to this vector, even though you had a little bit of overhead from converting in this direction and then converting back. So in a lot of problems, especially in computer science frankly, or some other applications you might be doing, you want to pick the right basis. The problems, or at least the computation for many problems, get a lot simpler if you pick the right coordinate system. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So let's say I have some matrix, let's call it matrix x. Matrix x is equal to, I'll just start with a 3 by 3 case because I think the 2 by 2 case is a bit trivial. Actually, why don't I just start with a 2 by 2 case? Let's say matrix x is a, b, and then it has x1, x2. I could have called these c and d, but you'll see why I call them x1 and x2 in a second. And now let's say I have another matrix. Let's say matrix y is identical to matrix x, except for this row. So matrix y is a, b, y1, and y2. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | I could have called these c and d, but you'll see why I call them x1 and x2 in a second. And now let's say I have another matrix. Let's say matrix y is identical to matrix x, except for this row. So matrix y is a, b, y1, and y2. And let's say we have a third matrix, z. Let's say we have a third matrix, z, that's identical to the first two matrices on the first row. So a, b. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So matrix y is a, b, y1, and y2. And let's say we have a third matrix, z. Let's say we have a third matrix, z, that's identical to the first two matrices on the first row. So a, b. But on the second row, it's actually the sum of the two rows of x and y. So this entry is going to be x1 plus y1. And this entry right here is y, sorry, is x2 plus y2. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So a, b. But on the second row, it's actually the sum of the two rows of x and y. So this entry is going to be x1 plus y1. And this entry right here is y, sorry, is x2 plus y2. Just like that. I want to be very clear. z is not x plus y. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | And this entry right here is y, sorry, is x2 plus y2. Just like that. I want to be very clear. z is not x plus y. All of the terms of z are not the sum of all the terms of x and y. I'm only focusing on one particular row. And this is just a general theme that you'll see over and over again, and we saw it in the last video, and I guess you'll see it here, is that the determinants, or finding the determinants, of matrices aren't linear on matrix operations, but they are linear on operations that you do just to one row. So in this case, everything else is equal except for this row, and z has the same first row as these guys, but its second row is the sum of the second row of these guys. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | z is not x plus y. All of the terms of z are not the sum of all the terms of x and y. I'm only focusing on one particular row. And this is just a general theme that you'll see over and over again, and we saw it in the last video, and I guess you'll see it here, is that the determinants, or finding the determinants, of matrices aren't linear on matrix operations, but they are linear on operations that you do just to one row. So in this case, everything else is equal except for this row, and z has the same first row as these guys, but its second row is the sum of the second row of these guys. So let's explore how the determinants of these guys relate. So the determinant, let me do it in x's color. The determinant of x, I'll write it like that, is equal to ax2 minus bx1. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So in this case, everything else is equal except for this row, and z has the same first row as these guys, but its second row is the sum of the second row of these guys. So let's explore how the determinants of these guys relate. So the determinant, let me do it in x's color. The determinant of x, I'll write it like that, is equal to ax2 minus bx1. We've seen that multiple times. The determinant of y is equal to ay2 minus by1. And the determinant of z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2, which is distributed to the a, minus bx1 minus by1. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | The determinant of x, I'll write it like that, is equal to ax2 minus bx1. We've seen that multiple times. The determinant of y is equal to ay2 minus by1. And the determinant of z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2, which is distributed to the a, minus bx1 minus by1. And if we just rearrange things, this is equal to a, let me write it this way, this is equal to ax2 minus bx1, so that's that term and that term. We switch colors. So that's those two guys. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | And the determinant of z is equal to a times x2 plus y2 minus b times x1 plus y1, which is equal to ax2 plus ay2, which is distributed to the a, minus bx1 minus by1. And if we just rearrange things, this is equal to a, let me write it this way, this is equal to ax2 minus bx1, so that's that term and that term. We switch colors. So that's those two guys. And then plus ay2 minus by1. Now, what is this right here? That is the determinant of x. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So that's those two guys. And then plus ay2 minus by1. Now, what is this right here? That is the determinant of x. And this right here is the determinant of y. So there you have it. If we have matrices that are completely identical, except for one row, and in this case it's a 2 by 2 matrix, so it looks like half of the matrix, and z's, that row that we're referring to that's different, z's is the sum of the other two guys' rows, then z's determinant is the sum of the other two determinants. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | That is the determinant of x. And this right here is the determinant of y. So there you have it. If we have matrices that are completely identical, except for one row, and in this case it's a 2 by 2 matrix, so it looks like half of the matrix, and z's, that row that we're referring to that's different, z's is the sum of the other two guys' rows, then z's determinant is the sum of the other two determinants. So this is a very special case. I want to keep reiterating, it only works in the case where this row and only this row is the sum of this row and this row, and the matrices are identical everywhere else. Let me show you the 3 by 3 case, and I think it'll make it a little bit more general. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | If we have matrices that are completely identical, except for one row, and in this case it's a 2 by 2 matrix, so it looks like half of the matrix, and z's, that row that we're referring to that's different, z's is the sum of the other two guys' rows, then z's determinant is the sum of the other two determinants. So this is a very special case. I want to keep reiterating, it only works in the case where this row and only this row is the sum of this row and this row, and the matrices are identical everywhere else. Let me show you the 3 by 3 case, and I think it'll make it a little bit more general. And then we'll go to n by n. The n by n is actually, on some level, the easiest to do, but it's kind of abstract, so I like to save that for the end. So let's redefine all those guys into the 3 by 3 case. So let's say that x is equal to a, b, c. Let's just make the third row the row we're going to use to determine our determinant. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | Let me show you the 3 by 3 case, and I think it'll make it a little bit more general. And then we'll go to n by n. The n by n is actually, on some level, the easiest to do, but it's kind of abstract, so I like to save that for the end. So let's redefine all those guys into the 3 by 3 case. So let's say that x is equal to a, b, c. Let's just make the third row the row we're going to use to determine our determinant. a, b, c, d, e, f. Actually, let me do the middle row, because I don't want to make you think it always has to be the last row. So let's say it's x1, x2, x3, and then you have a d, e, f. And what's the determinant of x going to be? The determinant of x is going to be equal to, let's say we're going along this row right here. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So let's say that x is equal to a, b, c. Let's just make the third row the row we're going to use to determine our determinant. a, b, c, d, e, f. Actually, let me do the middle row, because I don't want to make you think it always has to be the last row. So let's say it's x1, x2, x3, and then you have a d, e, f. And what's the determinant of x going to be? The determinant of x is going to be equal to, let's say we're going along this row right here. That's the row in question. It's going to be equal to, well, you remember your checkerboard pattern, so it's going to be, remember, plus, minus, plus, minus, plus. You remember all the rest, how it goes. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | The determinant of x is going to be equal to, let's say we're going along this row right here. That's the row in question. It's going to be equal to, well, you remember your checkerboard pattern, so it's going to be, remember, plus, minus, plus, minus, plus. You remember all the rest, how it goes. So it's going to start with a minus x1 times the submatrix. You get rid of that column, that row. b, c, e, f. b, c, e, f. Then you have plus x2 times the submatrix. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | You remember all the rest, how it goes. So it's going to start with a minus x1 times the submatrix. You get rid of that column, that row. b, c, e, f. b, c, e, f. Then you have plus x2 times the submatrix. Get rid of that column, that row. a, c, d, f. And then finally, minus x3. You get rid of its row and column. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | b, c, e, f. b, c, e, f. Then you have plus x2 times the submatrix. Get rid of that column, that row. a, c, d, f. And then finally, minus x3. You get rid of its row and column. You have a, b, d, e. Now, let me define another matrix, y, that is identical to matrix x, except for that row. So it's a, b, c, and then down here, d, e, f. But that middle row is different. It's y1, y2, and y3. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | You get rid of its row and column. You have a, b, d, e. Now, let me define another matrix, y, that is identical to matrix x, except for that row. So it's a, b, c, and then down here, d, e, f. But that middle row is different. It's y1, y2, and y3. What's the determinant of y going to be? Well, it's going to be identical to the determinant of x, because all the submatrices are going to be the same when you cross out this row and each of the columns. But the coefficients are going to be different. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | It's y1, y2, and y3. What's the determinant of y going to be? Well, it's going to be identical to the determinant of x, because all the submatrices are going to be the same when you cross out this row and each of the columns. But the coefficients are going to be different. Instead of an x1, you have a y1. So it's going to be equal to minus y1 times the determinant, b, c, e, f, plus y2 times the determinant of a, c, d, f, minus y3 times the determinant of a, b, d, e. I think you see where this is going. Now I'm going to create another matrix. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | But the coefficients are going to be different. Instead of an x1, you have a y1. So it's going to be equal to minus y1 times the determinant, b, c, e, f, plus y2 times the determinant of a, c, d, f, minus y3 times the determinant of a, b, d, e. I think you see where this is going. Now I'm going to create another matrix. I'm going to create another matrix, z. I'm going to create another matrix, z, just like that. That is equal to, it's identical to these two guys on the first and third rows, a, b, c, d, e, f, just like that. But this row just happens to be the sum of this row and this row. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | Now I'm going to create another matrix. I'm going to create another matrix, z. I'm going to create another matrix, z, just like that. That is equal to, it's identical to these two guys on the first and third rows, a, b, c, d, e, f, just like that. But this row just happens to be the sum of this row and this row. And when we figured out this determinant, we went along that row. You can see that right there. So this row right here is going to be x1 plus y1. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | But this row just happens to be the sum of this row and this row. And when we figured out this determinant, we went along that row. You can see that right there. So this row right here is going to be x1 plus y1. That's its first term. x2 plus y2. And then you have x3 plus y3. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So this row right here is going to be x1 plus y1. That's its first term. x2 plus y2. And then you have x3 plus y3. Now what's the determinant of z going to be? Well, we can go down this row right there. So it's going to be minus x1 plus y1 times its submatrix. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | And then you have x3 plus y3. Now what's the determinant of z going to be? Well, we can go down this row right there. So it's going to be minus x1 plus y1 times its submatrix. Get rid of that row, that column. You get b, c, e, f. You get b, c, e, f. I think you definitely see where this is going. Plus this coefficient, plus x2 plus y2 times its submatrix. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | So it's going to be minus x1 plus y1 times its submatrix. Get rid of that row, that column. You get b, c, e, f. You get b, c, e, f. I think you definitely see where this is going. Plus this coefficient, plus x2 plus y2 times its submatrix. Get rid of that row, that column. a, c, d, f. And then you have minus this guy right here. x3 plus y3 times its submatrix. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | Plus this coefficient, plus x2 plus y2 times its submatrix. Get rid of that row, that column. a, c, d, f. And then you have minus this guy right here. x3 plus y3 times its submatrix. Get rid of that column and row. a, b, d, e. Now what do you have right here? This is the determinant of z. |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | x3 plus y3 times its submatrix. Get rid of that column and row. a, b, d, e. Now what do you have right here? This is the determinant of z. This is, let me do it in a, this right here is the determinant of z. This thing right here. I think you can see immediately that if you were to add this to this, you would get this right here, right? |
Determinant when row is added Matrix transformations Linear Algebra Khan Academy.mp3 | This is the determinant of z. This is, let me do it in a, this right here is the determinant of z. This thing right here. I think you can see immediately that if you were to add this to this, you would get this right here, right? Because you would have this coefficient and this coefficient on that. If you added them up, you would get minus x1 plus y1. This guy and this guy add up to this guy. |
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