Math 1220 Linear Algebra

Example 7.4.7. Composition of linear operators.

• Let \(L_1\) be a rotation around the origin counterclockwise by an angle \(\phi\) and \(L_2\) be a rotation by an angle \(\theta\text{.}\) This means
  \begin{equation*} A_{L_1}= \begin{bmatrix} \cos \phi \amp -\sin\phi\\ \sin\phi \amp \cos\phi \end{bmatrix}\\ A_{L_2}= \begin{bmatrix} \cos \theta \amp -\sin\theta\\ \sin\theta \amp \cos\theta \end{bmatrix} \text{.} \end{equation*}
  Then \(L_1\circ L_2\) is a rotation by \(\theta\) followed by a rotation by an angle \(\phi\text{;}\) together it is a rotation through an angle of \(\theta+\phi\text{,}\) and so
  \begin{equation*} A_{L_1\circ L_2}= \begin{bmatrix} \cos(\theta+\phi) \amp -\sin(\theta+\phi)\\ \sin(\theta+\phi) \amp \cos(\theta+\phi) \end{bmatrix} \text{.} \end{equation*}
  It then follows from Theorem 7.4.4 that
  \begin{equation*} \begin{bmatrix} \cos(\theta+\phi) \amp -\sin(\theta+\phi)\\ \sin(\theta+\phi) \amp \cos(\theta+\phi) \end{bmatrix} = \begin{bmatrix} \cos \phi \amp -\sin\phi\\ \sin\phi \amp \cos\phi \end{bmatrix} \begin{bmatrix} \cos \theta \amp -\sin\theta\\ \sin\theta \amp \cos\theta \end{bmatrix} \text{.} \end{equation*}
  Computing the matrix product and equating the corresponding entries in the first column gives
  \begin{equation*} \cos(\theta+\phi) = \cos\theta \cos\phi-\sin\theta\sin\phi\\ \sin(\theta+\phi) = \cos\theta\sin\phi+\sin\theta\cos\phi\text{.} \end{equation*}
  This might be the world's shortest proof of the sum formulas for the sine and cosine functions.
• Let \(L\) be a reflection by the line \(y=x\text{,}\) and consider \(L^2\text{.}\) Then
  \begin{equation*} A_{L^2} =(A_L)^2 = \left(\begin{bmatrix} 0\amp1\\1\amp0 \end{bmatrix}\right)^2 =I\text{,} \end{equation*}
  and so \(L^2\) is the identity transformation. This is not surprising: the mirror image of a mirror image is just the original image. This can also be written as \(A=A^{-1}\text{.}\) A matrix satisfying this equation are called an involution.
• Let \(L\) be the projection onto the line \(y=x\text{,}\) and consider \(L^2\text{.}\) Then
  \begin{equation*} A_{L^2} =(A_L)^2 = \left(\frac12\begin{bmatrix} 1\amp1\\1\amp1 \end{bmatrix}\right)^2 =\frac14 \begin{bmatrix} 2\amp2\\2\amp2 \end{bmatrix} =\frac12 \begin{bmatrix} 1\amp1\\1\amp1 \end{bmatrix} =A_L\text{.} \end{equation*}
  A matrix \(A\) satisfying \(A^2=A\) is called idempotent.

Example 7.4.9.

Now we take another look at Figure 5.10.12 and Example 5.10.10 in which we wish to see that the graph of \(x^2-xy+y^2=1\) is an ellipse.

The new strategy is to rotate the graph clockwise through an angle of \(\frac\pi4\) and then verify that the new graph satisfies \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) for appropriate choice of \(a\) and \(b\text{.}\)

First we determine \(a\) and \(b\text{.}\) Since \((1,1)\) rotates into \((\sqrt2,0)\text{,}\) we must have \(a^2=2\text{.}\) Similarly, we must have \(b^2=\frac23\text{.}\) This means that \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) may be rewritten as \(x^2+3y^2=2\text{.}\)

Now suppose that \((x,y)\) is on the original graph, that is, \(x^2-xy+y^2=1\text{,}\) and suppose that \((x,y)\) rotates to \((u,v)\text{.}\) The matrix representation of the rotation is \(\frac1{\sqrt2} \left[\begin{smallmatrix} 1\amp1\\-1\amp1 \end{smallmatrix}\right] \text{,}\) and so

This implies

and so \((u,v)\) is on the ellipse with the equation \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}\)

Example 7.4.11.

In List 7.3.1 we saw that a reflection by the line \(y=x\) is a linear transformation. We now extend this to arbitrary lines through the origin with slope \(m\text{,}\) that is, lines with an equation of the form \(y=mx\text{.}\) Note that the point \((1,m)\) is on the line. If we let \(\theta\) be the angle between the line and the positive \(x\text{-axis}\text{,}\) then \(\cos\theta=\frac1{\sqrt{m^2+1}}\) and \(\sin\theta=\frac m{\sqrt{m^2+1}}\text{.}\)

By a clever use of composition of linear transformations, not only it can be seen that the reflection is a linear transformation, but also the actual formula can be revealed.

Consider the following sequence of linear transformations:

• \(L_1\text{:}\) Rotation clockwise by an angle \(\theta\text{,}\)
• \(L_2\text{:}\) Reflection by the \(x\)-axis,
• \(L_3\text{:}\) Rotation counterclockwise by an angle \(\theta\text{.}\)

Observe that the composition \(L_3\circ L_2\circ L_1\) has the effect: first \(L_1\text{,}\) the rotation of the plane clockwise by \(\theta\) rotates the line \(y=mx\) to the \(x\)-axis; \(L_2\text{,}\) the reflection by the \(x\)-axis, simply multiplies the second coordinate by \(-1\text{;}\) \(L_3\text{,}\) the rotation of the plane counterclockwise by \(\theta\) moves \(\mathbf u\) back to its original position. The combined effect is the reflection by the line \(y=mx\text{.}\)

The matrix representation for \(L_1\) is given in Checkpoint 7.3.19. Reflecting by the \(x\)-axis simply negates the second coordinate, and so the matrix representation for \(L_2\) is straightforward. Finally, Theorem 7.4.4 is used to compute the matrix representation of \(L_3\circ L_2\circ L_1\text{.}\) Recall that \(\cos\theta=\frac1{\sqrt{m^2+1}}\) and \(\sin\theta=\frac m{\sqrt{m^2+1}}\text{.}\)

It then follows that

Example 7.4.16.

Consider the following sequence of linear operators in \(\R^2\text{:}\)

1. \(L_1\text{:}\) reflect by the line \(y=x\text{,}\)
2. \(L_2\text{:}\) rotate counterclockwise by \(\theta=\frac\pi4\text{,}\)
3. \(L_3\text{:}\) reflect by the line \(y=-x\text{,}\)
4. \(L_4\text{:}\) rotate clockwise by \(\theta=\frac\pi4\text{,}\)
5. \(L_5\text{:}\) reflect by the line \(y=x\text{.}\)

The application of this sequence of operators results in an operator \(L=L_5\circ L_4\circ L_3\circ L_2\circ L_1\text{.}\) To understand the structure of \(L\text{,}\) look at the matrix representation in each case:

The matrix representation of the composition is then

Hence \(L((x,y))=(x,-y)\) and \(L\) is simply a reflection by the \(x\)-axis.