I'll get you a start here:

For the graphical part, start with, say, (3,7). Regard that as the coordinates in the standard X-Y coordinate system of a point on the plane. So, the X coordinate is the 3 and the Y coordinate is the 7. You could get out some graph paper and plot the thing. So, more generally, given two numbers x and y, (x,y) is the coordinates of a point in the plane. We call (x,y) a vector and imagine that it is an arrow from the origin to an arrow head at point (x,y). Then we can imagine that we can slide the vector around on the plane, keeping its length and direction the same.

What we did for the plane and X-Y we could do for space with X-Y-Z. So, there a vector would have three coordinates. Ah, call them components.

Now in linear algebra, for a positive integer n, we could have a vector with n components. For geometric intuition, what we saw in X-Y or X-Y-Z is usually enough.

We can let R denote the set of real numbers. Then R^n denotes the set of all of the vectors with n components that are real numbers. Our R^n is the leading example of a vector space. Sometimes it is good to permit the components to be complex numbers, but that is a little advanced. And the components could be elements of some goofy finite field from abstract algebra, but that also is a bit advanced.

Here we just stay with the real numbers, elements of R.

Okay, suppose someone tells us

ax + by = s

cx + dy = t

where the a, b, c, d, s, t are real numbers.

We want to know what the x and y are. Right, we can think of the vector (x,y). Without too much work we can show that, depending on the coefficients a, ..., t, the set of all (x,y), that fits the two equations has none, one, or infinitely many (points, vectors) solutions.

So, that example is two equations in two unknowns, x and y. Well, for positive integers m and n, we could have m equations in n unknowns. Still, there are none, one, or infinitely many solutions.

C. F. Gauss gave us Gauss elimination that lets us know if none, one, or infinitely many, find the one, or generate as many as we wish of the infinite.

We can multiply one of the equations by a number and add the resulting equation to one of the other equations. We just did an elementary row operation, and you can convince yourself that the set of all solutions remains the same. So, Gauss elimination is to pick elementary row operations that make the pattern of coefficients have a lot of zeros so that we can by inspection read off the none, one, or infinitely many. Gauss elimination is not difficult or tricky and programs easily in C, Fortran, ....

Quite generally in math, if we have a function f, some numbers a and b, and some things, of high generality, e.g., our vectors, and it is true that for any a, b and things x and y

f(ax + by) = af(x) + bf(y)

then we say that function f is linear. Now you know why our subject is called linear algebra.

A case of a linear function is Schroedinger's equation in quantum mechanics, and linear algebra can be a good first step into some of the math of quantum mechanics.

Let's see why those equations were linear: Let

f(x,y) = ax + by

Then

f[ c(x,y) + d(u,v)]

= f[ (cx, cy) + (du,dv) ]

= f(cx + du, cy + dv)

= a(cx + du) + b(cy + dv)

= c(ax) + d(au) + c(by) + d(bv)

= c(ax + by) + d(au + bv)

= cf(x,y) + df(u,v)

Done!

This linearity is mostly what makes linear algebra get its mathematical theorems and its utility in applications.

We commonly regard the plane with coordinates X-Y as 2 dimensional and space with coordinates X-Y-Z as 3 dimensional. If we study dimension carefully, then the 2 and 3 are correct. Similarly R^n is n dimensional.

We can write

ax + by = s

cx + dy = t

as x(a,c) + y(b,d) = (s,t)

So, (a,c), (b,d), and (s,t) are vectors, and x and y are coefficients that let us write vector (s,t) as a linear combination of the two vectors (a,c) and (b,d).

Apparently the superposition in quantum mechanics is closely related to this linear combination.

Well suppose these two vectors (a,c) and (b,d) can be used in such a linear combination to get any vector (s,t). Then, omitting some details, (a,c) and (b,d) span all of R^2, are linearly independent, and form a basis for the vector space R^2.

Sure, the usual basis for R^2 is just

(1,0)

(0,1)

And that our basis has two vectors is because R^2 is 2 dimensional. Works the same in R^n -- n dimensional and a basis has n vectors that are linearly independent.

Now for some geometric intuition,
given vectors in R^3 (x,y,z) and (u,v,w), then for coefficients a and b, the set of all

a(x,y,z) + b(u,v,w)

forms, depending on the two vectors, a point, a line, or a plane through (0,0,0) -- usually a plane and, thus, a vector subspace of dimension 0, 1, 2, usually 2.

And this works in R^n: We can have vector subspaces of dimension 0, 1, ..., n. For a subspace V of dimension m, 1 <= m <= n, there will be a basis of m linearly independent vectors in subspace V.

Let's explain matrix notation: Back to

ax + by = s

cx + dy = t

On the left side, let's rip out the x and y and write the rest as

/a b\
| |
\c d/

So, this matrix has two rows and two columns. Let's call this matrix A. For positive integers m and n, we can have a matrix with m rows and n columns and call it an m x n (pronounced m by n) matrix.

The (x,y) we can now call a 1 x 2 matrix. But we really want its transpose, 2 x 1 as

/x\
| |
\y/

Let's call this matrix v.

We want to define the matrix product

Av

We define it to be just what we saw in

ax + by = s

cx + dy = t

That is, Av is the transpose of (s,t).

If we have a vector u and coefficients a and b and define matrix addition in the obvious way, we can have

A(au + bv) = aAu + bAv

= a(Au) + b(Av)

= (aA)u + (bA)v

that is, we have some (associativity), so that A acts like a linear function. Right, the subject is linear algebra.

And matrix multiplication is associative, and the usual proof is just an application of the interchange of summation signs for finitely many terms.

We can define the length of a vector and the angle between two vectors. then multiplying two vectors by an orthogonal matrix U does not change the length or angle of two vectors.

Then for any orthogonal matrix U, all it does is reflect and/or make a rigid rotation.

We can also have a symmetric, positive definite matrix S. What S does is stretch a sphere into an ellipsoid (the 3 dimensional case does provide good intuition). Then A can be written as SU. That is, all A can do is rotate and reflect and then move a sphere into an ellipsoid. That is the polar decomposition and is the key to much of the most advanced work in linear algebra. Turns out, once we know more about orthogonal and symmetric matrices, the proof is short.

That's enough for a fast introduction!