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Course Notes for Linear Algebra

Remkes Kooistra

Section 6.6 Week 6 Activity

Subsection 6.6.1 Matrix Action

Activity 6.6.1.

In Subsection 6.3.2, I gave the matrices for rotations by \(\frac{\pi}{2}\) radians, \(\pi\) radians, and \(\frac{3\pi}{2}\) radians. I also defined reflections over the \(x\)-axis, \(y\)-axs, the line \(y=x\text{,}\) and the line \(y=-x\text{.}\) For all seven of these transformations, check that the matrix action acts as expected on the vectors \(\left( \begin{matrix} 1 \\ 0 \end{matrix} \right)\text{,}\) \(\left( \begin{matrix} 0 \\ 1 \end{matrix} \right)\text{,}\) \(\left( \begin{matrix} 1 \\ 1 \end{matrix} \right)\) and \(\left( \begin{matrix} 1 \\ -1 \end{matrix} \right)\text{.}\)
Solution.
I have seven matrices, each of which acts on the four given vectors. I’ll go through the actions one by one, starting with rotation by \(\frac{\pi}{2}\text{.}\)
\begin{align*} \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ 1 + 0 \end{matrix} \right) = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 - 1 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 - 1 \\ 1 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 1 \\ 1 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \end{align*}
All four vectors are indeed sent to their rotations by a quarter turn. Now look at the rotation by \(\pi\text{.}\)
\begin{align*} \left( \begin{matrix} -1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} -1 + 0 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} -1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ 0 - 1 \end{matrix} \right) = \left( \begin{matrix} 0 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} -1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -1 + 0 \\ 0 + -1 \end{matrix} \right) = \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} -1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} -1 + 0 \\ 0 + 1 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 1 \end{matrix} \right) \end{align*}
All of these vectors are indeed rotated by half a turn. (Notice that this is the same as “reflection” about the origin: each vector is sent to its negative. Now look at the rotation by \(\frac{3\pi}{2}\text{.}\)
\begin{align*} \left( \begin{matrix} 0 \amp 1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ -1 + 0 \end{matrix} \right) = \left( \begin{matrix} 0 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp 1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 1 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp 1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 1 \\ -1 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp 1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} 0 - 1 \\ -1 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) \end{align*}
All of these vectors are sent to their rotation by three quarters of a turn. Notice that I could also interpret this a rotation by a quarter turn clockwise. Now look at reflection over the \(x\)-axis.
\begin{align*} \left( \begin{matrix} 1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 1 + 0 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} 1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ 0 - 1 \\ \end{matrix} \right) = \left( \begin{matrix} 0 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 1 + 0 \\ 0 + -1 \end{matrix} \right) = \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} 1 + 0 \\ 0 + 1 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \end{align*}
The sign of the \(y\) coordinate of each vector is changed, which is exactly what I expect for reflection over the \(x\)-axis. Now look at reflection over the \(y\)-axis.
\begin{align*} \left( \begin{matrix} -1 \amp 0 \\ 0 \amp 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} -1 + 0 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} -1 \amp 0 \\ 0 \amp 1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ 0 + 1 \end{matrix} \right) = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} -1 \amp 0 \\ 0 \amp 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -1 + 0 \\ 0 + 1 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} -1 \amp 0 \\ 0 \amp 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} -1 + 0 \\ 0 - 1 \end{matrix} \right) = \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) \end{align*}
The \(x\) coordinate has now changed in sign, which is as expected. Now look at the reflection over the line \(y=x\text{.}\)
\begin{align*} \left( \begin{matrix} 0 \amp 1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ 1 + 0 \end{matrix} \right) = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp 1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 1 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp 1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 1 \\ 1 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp 1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} 0 - 1 \\ 1 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 1 \end{matrix} \right) \end{align*}
The two coordinates are exchanged, which is exactly what I would expect for this reflection. Finally, look at the reflection over the line \(y=-x\text{.}\)
\begin{align*} \left( \begin{matrix} 0 \amp -1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \left( \begin{matrix} 0 + 0 \\ -1 + 0 \end{matrix} \right) = \left( \begin{matrix} 0 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 - 1 \\ 0 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 0 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 0 - 1 \\ -1 + 0 \end{matrix} \right) = \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -1 \\ -1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} 0 + 1 \\ -1 + 0 \end{matrix} \right) = \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \end{align*}
The coordinates are switched and the signs are changed, which is what I would expect with this reflection.

Activity 6.6.2.

Calculate this matrix action.
\begin{equation*} \left( \begin{matrix} -5 \amp 0 \amp 3 \amp 2 \\ 4 \amp -5 \amp -7 \amp 1 \\ 1 \amp 2 \amp 0 \amp -8 \\ \end{matrix} \right) \left( \begin{matrix} 4 \\ -1 \\ -1 \\ -1 \end{matrix} \right) \end{equation*}
Solution.
I apply the matrix action algorithm: going along the rows of the matrix and down the column vector, multiplying paired terms and adding the products together.
\begin{align*} \amp \left( \begin{matrix} -5 \amp 0 \amp 3 \amp 2 \\ 4 \amp -5 \amp -7 \amp 1 \\ 1 \amp 2 \amp 0 \amp -8 \\ \end{matrix} \right) \left( \begin{matrix} 4 \\ -1 \\ -1 \\ -1 \end{matrix} \right) \\ \amp = \left( \begin{matrix} (-5)(4) + (0)(-1) + (3)(-1) + (2)(-1) \\ (4)(4) + (-5)(-1) + (-7)(-1) + (1)(-1) \\ (1)(4) + (2)(-1) + (0)(-1) + (-8)(-1) \end{matrix} \right) = \left( \begin{matrix} -25 \\ 27 \\ 10 \end{matrix} \right) \end{align*}

Activity 6.6.3.

Calculate this matrix action.
\begin{equation*} \left( \begin{matrix} -1 \amp 6 \amp 0 \amp -4 \\ 0 \amp 3 \amp 3 \amp -12 \\ -5 \amp -6 \amp -2 \amp 0 \\ 0 \amp -2 \amp -2 \amp 5 \end{matrix} \right) \left( \begin{matrix} 0 \\ 10 \\ -5 \\ -5 \end{matrix} \right) \end{equation*}
Solution.
I apply the matrix action algorithm: going along the rows of the matrix and down the column vector, multiplying paired terms and adding the products together.
\begin{align*} \amp \left( \begin{matrix} -1 \amp 6 \amp 0 \amp -4 \\ 0 \amp 3 \amp 3 \amp -12 \\ -5 \amp -6 \amp -2 \amp 0 \\ 0 \amp -2 \amp -2 \amp 5 \end{matrix} \right) \left( \begin{matrix} 0 \\ 10 \\ -5 \\ -5 \end{matrix} \right) \\ \amp = \left( \begin{matrix} (-1)(0) + (6)(10) + (0)(-5) + (-4)(-5) \\ (0)(0) + (3)(10) + (3)(-5) + (-12)(-5) \\ (-5)(0) + (-6)(10) + (-2)(-5) + (0)(-5) \\ (0)(0) + (-2)(10) + (-2)(-5) + (5)(-5) \\ \end{matrix} \right) = \left( \begin{matrix} 80 \\ 75 \\ -50 \\ -35 \end{matrix} \right) \end{align*}

Activity 6.6.4.

Calculate this matrix action.
\begin{equation*} \left( \begin{matrix} 5 \amp -3 \amp 9 \amp 0 \amp 2 \amp -7 \\ 2 \amp -2 \amp 0 \amp -5 \amp -6 \amp -9 \end{matrix} \right) \left( \begin{matrix} 0 \\ -3 \\ -3 \\ 7 \\ 2 \\ 4 \end{matrix} \right) \end{equation*}
Solution.
I apply the matrix action algorithm: going along the rows of the matrix and down the column vector, multiplying paired terms and adding the products together.
\begin{align*} \amp \left( \begin{matrix} 5 \amp -3 \amp 9 \amp 0 \amp 2 \amp -7 \\ 2 \amp -2 \amp 0 \amp -5 \amp -6 \amp -9 \end{matrix} \right) \left( \begin{matrix} 0 \\ -3 \\ -3 \\ 7 \\ 2 \\ 4 \end{matrix} \right)\\ \amp = \left( \begin{matrix} (5)(0) + (-3)(-3) + (9)(-3) + (0)(7) + (2)(2) + (-7)(4) \\ (2)(0) + (-2)(-3) + (0)(-3) + (-5)(7) + (-6)(2) + (-9)(4) \end{matrix} \right) \\ \amp = \left( \begin{matrix} 0 + 9 - 27 + 0 + 4 - 28 \\ 0 + 6 + 0 - 35 - 12 - 36 \end{matrix} \right) = \left( \begin{matrix} -42 \\ -77 \end{matrix} \right) \end{align*}

Activity 6.6.5.

Calculate this matrix action.
\begin{equation*} \left( \begin{matrix} -5 \amp 3 \amp 0 \\ 2 \amp 0 \amp -4 \\ -10 \amp 1 \amp 1 \\ -13 \amp -2 \amp 5 \\ 5 \amp 4 \amp -9 \end{matrix} \right) \left( \begin{matrix} 1 \\ -2 \\ 0 \end{matrix} \right) \end{equation*}
Solution.
I apply the matrix action algorithm: going along the rows of the matrix and down the column vector, multiplying paired terms and adding the products together.
\begin{align*} \amp \left( \begin{matrix} -5 \amp 3 \amp 0 \\ 2 \amp 0 \amp -4 \\ -10 \amp 1 \amp 1 \\ -13 \amp -2 \amp 5 \\ 5 \amp 4 \amp -9 \end{matrix} \right) \left( \begin{matrix} 1 \\ -2 \\ 0 \end{matrix} \right) \\ \amp = \left( \begin{matrix} (-5)(1) + (3)(-2) + (0)(0) \\ (2)(1) + (0)(-2) + (-4)(0) \\ (-10)(1) + (1)(-2) + (1)(0) \\ (-13)(1) + (-2)(-2) + (5)(0) \\ (5)(1) + (4)(-2) + (-9)(0) \\ \end{matrix} \right) = \left( \begin{matrix} -11 \\ 2 \\ -12 \\ -9 \\ -3 \end{matrix} \right) \end{align*}

Subsection 6.6.2 Matrix Multiplication

Activity 6.6.6.

Multiply these two matrices.
\begin{equation*} \left( \begin{matrix} 9 \amp -1 \amp -1 \\ 0 \amp 5 \amp -5 \\ -3 \amp -3 \amp -3 \end{matrix} \right) \left( \begin{matrix} 0 \amp -2 \amp -8 \\ 1 \amp 1 \amp -8 \\ 9 \amp 0 \amp 3 \end{matrix} \right) \end{equation*}
Solution.
I use the algorithm, going across the rows of the first matrix and down the rows of the second.
\begin{equation*} \left( \begin{matrix} 9 \amp -1 \amp -1 \\ 0 \amp 5 \amp -5 \\ -3 \amp -3 \amp -3 \end{matrix} \right) \left( \begin{matrix} 0 \amp -2 \amp -8 \\ 1 \amp 1 \amp -8 \\ 9 \amp 0 \amp 3 \end{matrix} \right) = \left( \begin{matrix} -10 \amp -19 \amp -67 \\ -40 \amp 5 \amp -55 \\ -30 \amp 3 \amp 39 \end{matrix} \right) \end{equation*}

Activity 6.6.7.

Multiply these two matrices.
\begin{equation*} \left( \begin{matrix} 7 \amp -1 \amp -1 \amp 3 \\ 5 \amp -9 \amp 0 \amp 1 \end{matrix} \right) \left( \begin{matrix} 4 \amp -4 \\ 1 \amp 0 \\ -3 \amp -3 \\ 1 \amp 5 \end{matrix} \right) \end{equation*}
Solution.
I use the algorithm, going across the rows of the first matrix and down the rows of the second.
\begin{equation*} \left( \begin{matrix} 7 \amp -1 \amp -1 \amp 3 \\ 5 \amp -9 \amp 0 \amp 1 \end{matrix} \right) \left( \begin{matrix} 4 \amp -4 \\ 1 \amp 0 \\ -3 \amp -3 \\ 1 \amp 5 \end{matrix} \right) = \left( \begin{matrix} 33 \amp -10 \\ 12 \amp -15 \end{matrix} \right) \end{equation*}

Activity 6.6.8.

Take the matrix for rotation by \(\frac{\pi}{2}\) radians in \(\RR^2\) and muliply it by itself to check that the result is the matrix for rotation by \(\pi\) radians. Multiply by the original matrix again and check that the result is now the matrix for rotation by \(\frac{3\pi}{2}\text{.}\)
Solution.
The matrix for a quarter turn rotation is
\begin{equation*} \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) \end{equation*}
To compose transformations we multiply matrices. To compose this transformation with itself, I mutiply the matrix with itself.
\begin{equation*} \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) = \left( \begin{matrix} -1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \end{equation*}
The result of the matrix multiplication is indeed the rotation matrix for rotation by \(\pi\) radians. To see what happens after a third rotation by a quarter turn, I multiply this matrix by the original quarter turn matrix.
\begin{equation*} \left( \begin{matrix} -1 \amp 0 \\ 0 \amp -1 \end{matrix} \right) \left( \begin{matrix} 0 \amp -1 \\ 1 \amp 0 \end{matrix} \right) = \left( \begin{matrix} 0 \amp 1 \\ -1 \amp 0 \end{matrix} \right) \end{equation*}
The result of the matrix multiplication is indeed the matrix of the rotation by \(\frac{3\pi}{2}\) radians.

Subsection 6.6.3 Proof Questions

Activity 6.6.9.

Consider two rotations in \(\RR^2\text{,}\) by angles \(\theta\) and \(\phi\text{.}\) Using the general form of a matrix for a rotation, prove that the composition of these two rotations is the rotation by the angle \((\theta + \phi)\text{.}\) (You will need to use some trigonometric identities to prove this.)
Solution.
The general form of a rotation by \(\theta\) is
\begin{equation*} \left( \begin{matrix} \cos \theta \amp - \sin \theta \\ \sin \theta \amp \cos \theta \end{matrix} \right) \text{.} \end{equation*}
I can make the same matrix for the angle \(\phi\text{.}\)
\begin{equation*} \left( \begin{matrix} \cos \phi \amp - \sin \phi \\ \sin \phi \amp \cos \phi \end{matrix} \right) \text{.} \end{equation*}
I compose these two matrices.
\begin{align*} \amp \left( \begin{matrix} \cos \theta \amp - \sin \theta \\ \sin \theta \amp \cos \theta \end{matrix} \right) \left( \begin{matrix} \cos \phi \amp - \sin \phi \\ \sin \phi \amp \cos \phi \end{matrix} \right)\\ \amp = \left( \begin{matrix} (\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{matrix} \right) \end{align*}
Now I need trig identities. Each of these four matrix entries is a sum identity. Replacing those according to identities drawn from the reference material, I get this matrix.
\begin{equation*} \left( \begin{matrix} \cos (\theta + \phi) \amp -\sin (\theta + \phi) \\ \sin (\theta + \phi) \amp \cos (\theta + \phi) \end{matrix} \right) \end{equation*}
This is precisely the matrix form for rotation by the angle \((\theta + \phi)\text{.}\) Therefore, I have proven that the matrix multiplication, and hence the composition, of two rotations produces the rotation by the sum of the two angles.

Activity 6.6.10.

Prove that any reflection in \(\RR^2\) composes with itself to produce the identity. (Use the general matrix for these reflections).
Solution.
The general form for a reflection over the line spanned by the unit vector \(\left( \begin{matrix} a \\ b \end{matrix} \right)\) is
\begin{equation*} \left( \begin{matrix} a^2 - b^2 \amp 2ab \\ 2ab \amp b^2 - a^2 \end{matrix} \right) \text{.} \end{equation*}
Since the vector defining the line is a unit vector, I know \(a^2 + b^2 = 1\text{.}\) This fact will be important to the proof. Now I take this matrix and multiply it by itself to do the composition of the reflection with itself. I’ll have to do some algebra with the resulting terms in the new matrix.
\begin{align*} \amp \left( \begin{matrix} a^2 - b^2 \amp 2ab \\ 2ab \amp b^2 - a^2 \end{matrix} \right) \left( \begin{matrix} a^2 - b^2 \amp 2ab \\ 2ab \amp b^2 - a^2 \end{matrix} \right) \\ \amp = \left( \begin{matrix} (a^2 - b^2)^2 + (2ab)(2ab) \amp (a^2-b^2)(2ab) + (2ab)(b^2-a^2) \\ (2ab)(a^2-b^2) + (b^2-a^2)(2ab) \amp (2ab)(2ab) + (b^2-a^2)^2 \end{matrix} \right) \\ \amp = \left( \begin{matrix} a^4 - 2a^2b^2 + b^4 + 4a^2b^2 \amp (a^2-b^2)(2ab) - (a^2-b^2)(2ab) \\ (2ab)(a^2-b^2) - (2ab)(a^2-b^2) \amp 4a^2b^2 + b^4 - 2a^2b^2 + a^4 \end{matrix} \right) \\ \amp = \left( \begin{matrix} a^4 + 2a^2b^2 + b^4 \amp 0 \\ 0 \amp b^2 + 2a^2b^2 + a^2 \end{matrix} \right) \\ \amp = \left( \begin{matrix} (a^2+b^2)^2 \amp 0 \\ 0 \amp (a^2+b^2)^2 \end{matrix} \right) \\ \amp = \left( \begin{matrix} 1 \amp 0 \\ 0 \amp 1 \end{matrix} \right) \end{align*}
In the last step, I use the important fact about the unit vector. I end up with the identity matrix, which is what I wanted, so the proof is finished.

Subsection 6.6.4 Conceptual Review Questions

  • Why can matrices represent linear transformation?
  • What is the matrix action?
  • What is an associative binary operations? What is a commutative binary operation?
  • Why isn’t matrix multiplication commutative?
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