Section 6.5 Proofs - Matrix Actions
Subsection 6.5.1 General Properties of the Matrix Action
I have defined linear transformations. At this point, I could state and try to prove a variety of properties about linear transformations and their matrix representations. Most obviously, perhaps, I could prove that the matrix action is actually a linear transformation, as I stated in Section 6.2, and the converse that all linear transformations can be represented as matrices. I could also prove the two properties in Proposition 6.2.7. I am opting not to present these general proofs. They are not conceptually difficult, but to do them, in general, is notationally difficult. It involves a lot of matrix coefficients, vector coefficients and many subscripts. To include those proofs wasn’t useful for building up experience with proofs (other than perhaps to show that notation is a major challenge in writing mathematical proof).
Subsection 6.5.2 Properties of Transformations in \(\RR^2\)
Instead of proving the most general statements about matrix transformations, I’ll prove some more specific statements about the types of transformations I defined in Section 6.3.
Proof.
To commute means that if the order of the composition (or matrix multiplication) is reversed, the outcome is the same. That is, if \(A\) and \(B\) are the matrix representations of two rotations, then \(AB = BA\) is a true statement about matrix multiplication.
I want my proof to be completely general, so I want to work with arbitrary rotations. Let’s say that \(A\) is a rotation by the angle \(\theta\text{,}\) so it has the matrix
\begin{equation*}
\begin{pmatrix}
\cos \theta \amp - \sin \theta \\
\sin \theta \amp \cos \theta
\end{pmatrix}\text{.}
\end{equation*}
Let’s also say that \(B\) is a rotation by the angle \(\phi\text{,}\) so it has the matrix
\begin{equation*}
\begin{pmatrix}
\cos \phi \amp - \sin \phi \\
\sin \phi \amp \cos \phi
\end{pmatrix}\text{.}
\end{equation*}
Now I need to calculate the two matrix multiplications and show that the result is the same regardless of the order. First I will calculate \(AB\text{.}\)
\begin{align*}
AB \amp = \begin{pmatrix}
\cos \theta \amp - \sin \theta \\
\sin \theta \amp \cos \theta
\end{pmatrix}
\begin{pmatrix}
\cos \phi \amp - \sin \phi \\
\sin \phi \amp \cos \phi
\end{pmatrix} \\
\amp = \begin{pmatrix}
\cos \theta \cos \phi - \sin \theta \sin \phi \amp
-\cos \theta \sin \phi - \sin \theta \cos \phi \\
\sin \theta \cos \phi + \cos \theta \sin \phi \amp
-\sin \theta \sin \phi + \cos \theta \cos \phi
\end{pmatrix} \text{.}
\end{align*}
Now I’ll do the other order of the multiplication.
\begin{align*}
BA \amp = \begin{pmatrix}
\cos \phi \amp - \sin \phi \\
\sin \phi \amp \cos \phi
\end{pmatrix}
\begin{pmatrix}
\cos \theta \amp - \sin \theta \\
\sin \theta \amp \cos \theta
\end{pmatrix} \\
\amp = \begin{pmatrix}
\cos \phi \cos \theta - \sin \phi \sin \theta \amp
- \cos \phi \sin \theta - \sin \phi \cos \theta \\
\sin \phi \cos \theta + \cos \phi \sin \theta \amp
-\sin \phi \sin \theta + \cos \phi \cos \theta
\end{pmatrix} \text{.}
\end{align*}
I can take this second result and switch the order of all the multiplications (which is valid since these are just multiplications of ordinary numbers). I can also reorder a couple of the additions/subtractions.
\begin{equation*}
\begin{pmatrix}
\cos \theta \cos \phi - \sin \theta \sin \phi \amp
- \cos \theta \sin \phi - \sin \theta \cos \phi \\
\sin \theta \cos \phi + \cos \theta \sin \phi \amp
-\sin \theta \sin \phi + \cos \theta \cos \phi
\end{pmatrix} \text{.}
\end{equation*}
This is the same as the result of \(AB\text{.}\) This concludes the proof, showing that the outcome of either order of matrix multiplication is the same.
You might wonder, at this point, about the structure of this proof. I’ve tried to say several times that to prove an equality, I need to start at one side and end at the other. However, here I’ve tried to prove \(AB=BA\) and done so by reducing to a common form. Am I disobeying my own rules?
I am not contradicting myself, though it is a very good question. To understand why, look at the statement of the proposition. The composition of the transformation is a process where two transformations combine into one new one. There are two ways to do this: \(AB\) or \(BA\text{.}\) The proposition says that the outcome of both orders is the same. Therefore, I can calculate the outcome of both orders and compare them. This is different from a proposition that simply states an equation; for these propositions, reducing each side of the equation to a common expression can result in a false proof (for example, by multiplying both sides by zero). Here, I’m not reducing an equation to a common form but doing two processes and comparing the result.
Proposition 6.5.2.
Prove that a projection in \(\RR^2\) composed with itself is just the same projection. (That is, for any projection \(A\text{,}\) \(AA = A\text{.}\))Proof.
\begin{equation*}
A = \begin{pmatrix}
a^2 \amp ab \\
ab \amp b^2
\end{pmatrix} \text{,}
\end{equation*}
remembering that the vector defining the line is a unit vector direction. Since it is a unit vector direction, I know that \(a^2 + b^2 = 1\text{,}\) which I will need for this proof. Now I’ll take this matrix and multiply it by itself.
\begin{equation*}
A^2 = \begin{pmatrix}
a^2 \amp ab \\
ab \amp b^2
\end{pmatrix}
\begin{pmatrix}
a^2 \amp ab \\
ab \amp b^2
\end{pmatrix} =
\begin{pmatrix}
(a^2)(a^2) + (ab)(ab) \amp
(a^2)(ab) + (ab)(b^2) \\
(ab)(a^2) + (b^2)(ab) \amp
(ab)(ab) + (b^2)(b^2)
\end{pmatrix}
\end{equation*}
In each component of this matrix, I can factor out \(a^2\text{,}\) \(b^2\text{,}\) or \(ab\text{.}\)
\begin{equation*}
= \begin{pmatrix}
a^2(a^2 + b^2) \amp
ab(a^2 + b^2) \\
ab(a^2 + b^2) \amp
b^2(a^2 + b^2)
\end{pmatrix}
\end{equation*}
Then I can use the fact that \(a^2 + b^2 = 1\text{.}\)
\begin{equation*}
= \begin{pmatrix}
a^2 \amp ab \\
ab \amp b^2
\end{pmatrix} = A
\end{equation*}
This is the same as the starting matrix. I’ve proved that \(AA=A\text{,}\) which was the goal of the proposition.
