## Linear Algebra and Its Applications, Exercise 2.6.12

Exercise 2.6.12. If $H$ is the reflection matrix in the $x$ $y$ plane, show that $H^2 = I$ using the trigonometric identity $\cos^2 \theta + \sin^2\theta = 1$ ( $c^2+s^2=1$ for short). $H = \begin{bmatrix} 2c^2 -1&2cs \\ 2cs&2s^2-1 \end{bmatrix}$

so that $H^2 = \begin{bmatrix} 2c^2 -1&2cs \\ 2cs&2s^2-1 \end{bmatrix} \begin{bmatrix} 2c^2 -1&2cs \\ 2cs&2s^2-1 \end{bmatrix}$ $= \begin{bmatrix} (2c^2 -1)(2c^2 -1)+(2cs)(2cs)&2cs(2c^2-1)+2cs(2s^2-1) \\ 2cs(2c^2-1)+2cs(2s^2-1)&(2cs)(2cs)+(2s^2 -1)(2s^2 -1) \end{bmatrix}$ $= \begin{bmatrix} 4c^4-4c^2+1+4c^2s^2&2cs(2c^2+2s^2-2) \\ 2cs(2c^2+2s^2-2)&4c^2s^2+4s^4-4s^2+1 \end{bmatrix}$ $= \begin{bmatrix} 4c^2(c^2+s^2-1)+1&4cs(c^2+s^2-1) \\ 4cs(c^2+s^2-1)&4s^2(c^2+s^2-1)+1 \end{bmatrix}$

Since $c^2+s^2 =1$ this can be simplified to $H^2 = \begin{bmatrix} 4c^2(1-1)+1&4cs(1-1) \\ 4cs(1-1)&4s^2(1-1)+1 \end{bmatrix}$ $= \begin{bmatrix} 4c^2 \cdot 0+1&4cs \cdot 0 \\ 4cs \cdot 0&4s^2 \cdot 0+1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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