The first thing to know, which may be obvious, is that a vector is analogous to a column of a matrix. If we want to talk about a row of a matrix, we need to transpose the vector. I am using an apostrophe, , as the transpose operator. For example:

is a column vector, while is a row vector. (A lot of times, I will get lazy and write a column vector as a row vector to save time, but the meaning is ALWAYS supposed to be what I just wrote here)

This distinction seems stupid right now, but it becomes very important later. Always keep this in the back of your head. The dimensions of arrays and matrices matter a lot when things get more complicated, even though it seems trivially obvious right now.

So, on to matrix operations. Recall our matrix:

Suppose we want to multiply this matrix by a single number. We will call this number for now, and we write this like this:

Let’s see what is going on here:

That can be thought of as a 1×1 matrix, and we call that a constant. When we multiply a constant by a matrix, each element in the matrix gets multiplied by that constant.

Typically, we can only add matrices that have the same dimensions. When we do this, we add element by element:

Here, the two matrices are both 2×2. We can add matrices as long as they both have the same dimensions; such as 2×3, 4×7, lxw, etc.

Finally, let’s quickly look at matrix multiplication. Again, the dimensions matter. The rule of thumb for multiplying matrices is that the first matrix has to have the same number of columns as the second matrix has rows. For example, if we have one matrix that is 3×4, then it can be multiplied by any matrix with 4 rows: 4×3, 4×1, 4×4646. The order of the multiplication matters here! It is not always true that ! This should be obvious if you think about matrices that are not square.

I am being vague here, because I don’t want to get into too many of the details. Here is an example of how this works, because for now I only want to emphasize the mechanics of it:

A 2×3 times a 3×2 gives us a 2×2. This is true in general: an axb times a bxc gives us an axc. The general idea here is that you take the first column of the matrix and multiply it by each row of in turn. This populates the first column of the answer. Then you take the next column of and repeat to populate the next column of the answer.

Let’s go back to our linear equation:

There must be an easier way to write that. In fact, I would guess that originally linear algebra started as a way to write math more compactly, and then the cleaner view of the math allowed for more insight and the development of the theory.

First, let’s talk about vectors. Vectors are just lists that can be represented as a single variable name. For example, in our linear equation above, let’s define some vectors:

and

This could also be written as:

and

These are the same thing! I sort of prefer the first way, with the vectors being BOLD letters, because if I write this by hand I hate to put the little arrow over the letter. From here on out, I’m using the first, BOLD, notation for vectors.

Before we go too far with vectors, let’s talk about matrices, too. Here is a matrix:

The only thing I want to point out for right now is that we use a BOLD CAPITAL letter for the matrix, and we can identify an element in the matrix by counting down the rows first and then into the columns. The element is in ROW 2, COLUMN 3.

Back to our linear equation, suppose it were possible to write this:

instead as this:

It is possible, and in fact that is how you do it! More on the notation later!

Suppose we know that a solution to this equation is . In this case, this means that . This is the definition that most books stop at, because it seems very obvious what is going on here. A note I want to stress is that this is an equality, not a formula — from here on out, you can put anywhere you see , and vice versa.

The thing that I want to push a little more on, though, is this: Suppose we have another solution to this equation . This would mean that also.

So we have two equations, both of which equal . Let’s look at what happens when we set them equal to each other:

(subtract the right side)

(use distributive property)

Now we see that the solution is a solution to the equation

These two equations are clearly related. For now, just keep in mind the coefficients for both of these equations were the same, and the only difference was the right side.

Any way you look at it, I hate these books. I just want to learn the stuff. And so I present ClearMath. I hope to pick a different subject for each blog post that I have found horribly described in every textbook I’ve looked it up in, and do it a <hopefully> better justice here. At the very least, if anyone reads this blog, they can tell me that I have no idea what I am talking about or that I am not clear at all. And then we can fix it. Textbooks are so stale; they need some interaction to be useful!