Matrix Algebra from a Geometric Point of View

Suppose that \(f : \mathbb{R}^2 \to \mathbb{R}^3\) is defined by \(f\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+y\\y\\-y\end{bmatrix}\) and \(g : \mathbb{R}^3 \to \mathbb{R}^3\) is defined by \(g\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right) = \begin{bmatrix}x\\x+y\\x+y+z\end{bmatrix}\text{.}\) Then \(g \circ f : \mathbb{R}^2 \to \mathbb{R}^3\) is given by

In this example the composition in the other order, \(f \circ g\text{,}\) does not make sense. This is because the output of \(g\) is in \(\mathbb{R}^3\text{,}\) but \(f\) accepts inputs from \(\mathbb{R}^2\text{,}\) so the expression \(f\left(g\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right)\right) = f\left(\begin{bmatrix}x\\x+y\\x+y+z\end{bmatrix}\right)\) is meaningless.

Let \(f : \mathbb{R}^2 \to \mathbb{R}^2\) and \(g : \mathbb{R}^2 \to \mathbb{R}^2\) be defined by \(f\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+y\\x-y\end{bmatrix}\) and \(g\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}y\\2x\end{bmatrix}\text{.}\) In this case both \(g \circ f\) and \(f \circ g\) are defined. If we calculate them, we get:

and

As we can see, the formulas are different.

To be absolutely certain they they are really different (so that there is no clever way of re-writing the formulas to make them the same) we should pick a specific vector to plug into both and see that we get different answers. There are lots of different choices here, so we just have to pick one. For instance, \((g \circ f)\left(\begin{bmatrix}1\\0\end{bmatrix}\right) = \begin{bmatrix}1\\2\end{bmatrix}\) and \((f \circ g)\left(\begin{bmatrix}1\\0\end{bmatrix}\right) = \begin{bmatrix}2\\-2\end{bmatrix}\text{.}\)

One final caution: There are some choices of vector where \(g\circ f\) and \(f\circ g\) do give the same answer (for instance, this happens with \(\begin{bmatrix}0\\0\end{bmatrix}\)). That's OK, but we still have \(g \circ f \neq f \circ g\) because they disagree on at least one input.

Let \(T : \mathbb{R}^2 \to \mathbb{R}^2\) be the transformation that takes a vector \(\vec{v}\text{,}\) rotates it counterclockwise by \(\pi/6\text{,}\) then doubles the result of that, then takes the result of that and performs the orthogonal projection in the direction of \(\begin{bmatrix}3\\-1\end{bmatrix}\text{.}\) Then we can write \(T = \proj_{\begin{bmatrix}3\\-1\end{bmatrix}} \circ S \circ R_{\pi/6}\text{,}\) where \(S(\vec{v}) = 2\vec{v}\) and \(R_{\pi/6}\) is the rotation counterclockwise by \(\pi/6\text{.}\) Since all three of the terms in the composition are things that we already know are linear transformations, it follows from Theorem 4.3.5 that \(T\) is also a linear transformation.

Notice that using Theorem 4.3.5, together with some functions that we already know are linear transformations, has saved us from a fairly unpleaseant direct verification of the properties in Definition 4.1.2!

Let \(T : \mathbb{R}^2 \to \mathbb{R}^3\) and \(S : \mathbb{R}^3 \to \mathbb{R}^4\) be defined by \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+2y \\ 2x+y \\ x-y\end{bmatrix}\) and \(S\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right) = \begin{bmatrix}x-y+z \\ x+y \\ y-z \\ 2y+3z\end{bmatrix}\text{.}\) Then the composition \(S \circ T\) makes sense, and we calculate a formula for it:

Now if we compute the standard matrices (Definition 4.1.8), we find:

It is far from obvious how to combine \([T]\) and \([S]\) to obtain \([S \circ T]\text{!}\)

Let \(A = \begin{bmatrix}1 \amp 2 \\ 3 \amp 4 \\ 5 \amp 6\end{bmatrix}\) and \(B = \begin{bmatrix}0 \amp 1 \\ 2 \amp 3\end{bmatrix}\text{.}\) Then the first column of \(AB\) is

and the second column is

Therefore

We return to the transformations from Example 4.3.7. In that example we did the composition by hand, and then used the definition of the standard matrix to find:

With Theorem 4.3.10 we can now get the same answer for \([S \circ T]\) without having to actually perform the composition, because

Even better, from here we can get back the formula for the composition if we want it:

Fix two angles \(\theta\) and \(\phi\text{,}\) and let \(R_\theta, R_\phi : \mathbb{R}^2 \to \mathbb{R}^2\) be the counterclockwise rotations by \(\theta\) and \(\phi\text{,}\) respectively. Geometrically, you can see that rotating a vector first by \(\theta\) and then by \(\phi\) is the same as rotating the vector by \(\phi+\theta\text{,}\) and so \(R_\phi \circ R_\theta = R_{\phi +\theta}\text{.}\) By Theorem 4.3.10 we therefore have

In Example 4.1.15 we calculated the standard matrix for a rotation, so we can write out both sides of the above equation explicitly:

Now if we calculate the matrix multiplication on the left side, we get:

For these two matrices to be equal they must have equal corresponding entries. In particular, if we look at the top left entries we recover the addition formula for cosine, and if we look at the bottom left entries we get the addition formula for sine:

Notice how little trigonometry we needed in order to get these identities! All we used was the calculations in Example 4.1.15, the fact that the composition of two rotations is the rotation by the sum of the angles, and some very general facts about composition of linear transformations.

If \(A = \begin{bmatrix}1 \amp 2 \\ 3 \amp 4\end{bmatrix}\) and \(B = \begin{bmatrix}1 \amp 1 \\ 2 \amp 0\end{bmatrix}\) then \(AB = \begin{bmatrix}5 \amp 1 \\ 11 \amp 3\end{bmatrix}\) and \(BA = \begin{bmatrix}4 \amp 6 \\ 2 \amp 4\end{bmatrix}\text{.}\)

Let \(A = \begin{bmatrix}0 \amp 1 \\ 0 \amp 0\end{bmatrix}\) and \(B = \begin{bmatrix}0 \amp 0 \\ 0 \amp 1\end{bmatrix}\text{.}\) Notice that \(A \neq 0_{2 \times 2}\) and \(B \neq 0_{2 \times 2}\text{,}\) yet \(BA = 0_{2 \times 2}\text{!}\) Perhaps even more surprisingly, \(AB = \begin{bmatrix}0 \amp 1 \\ 0 \amp 0\end{bmatrix} \neq 0_{2 \times 2}\text{.}\)

It is a good exercise to find the linear transformations with matrices \(A\) and \(B\text{,}\) and then try to understand geometrically what the calculations in this example mean.

Let \(A = \begin{bmatrix}1 \amp 0 \\ 0 \amp 1\end{bmatrix}\text{,}\) \(B = \begin{bmatrix}-1 \amp 0 \\ 0 \amp -1\end{bmatrix}\text{,}\) \(C = \begin{bmatrix}-1 \amp 0 \\ 0 \amp 1\end{bmatrix}\text{,}\) and \(D = \begin{bmatrix}2 \amp 3 \\ -1 \amp -2\end{bmatrix}\text{.}\) Then

This example shows that the equation \(X^2 = I_2\text{,}\) where \(X\) is a \(2 \times 2\) matrix, has more than two solutions. It turns out that this equation actually has infinitely many solutions! Compare this with the situation of the equation \(x^2=1\) when \(x\) is a number.

Even if a matrix has no entries that are \(0\) products with that matrix can still yield the zero matrix. For example, \(\begin{bmatrix}-2 \amp 1 \\ -4 \amp 2\end{bmatrix}^2 = \begin{bmatrix}0 \amp 0 \\ 0\amp 0\end{bmatrix}\text{.}\)

Let \(A = \begin{bmatrix}2 \amp 1 \\ -1 \amp 0 \\ 3 \amp 2\end{bmatrix}\) and \(B = \begin{bmatrix}1 \amp 2 \\ 3 \amp 4\end{bmatrix}\text{.}\) Then

so

On the other hand,

In this example the product \(A^tB^t\) is not even defined. Even in examples where \(A^tB^t\) is defined, we often have \((AB)^t \neq A^tB^t\text{.}\)