1.

1. (a) T

2. (b) F

3. (c) F

4. (d) T

5. (e) F

6. (f) T

7. (g) F

8. (h) F

9. (i) F

2.

1. (a) $P={\displaystyle \frac{1}{\sqrt{2}}}\left(\begin{array}{rr}1& 1\\ 1& -1\end{array}\right)\text{and}D=\left(\begin{array}{rr}3& 0\\ 0& -1\end{array}\right)$

2. (d) $P=\left(\begin{array}{rrr}{\displaystyle \frac{1}{\sqrt{2}}}& {\displaystyle \frac{1}{\sqrt{6}}}& {\displaystyle \frac{1}{\sqrt{3}}}\\ -{\displaystyle \frac{1}{\sqrt{2}}}& {\displaystyle \frac{1}{\sqrt{6}}}& {\displaystyle \frac{1}{\sqrt{3}}}\\ 0& -{\displaystyle \frac{2}{\sqrt{6}}}& {\displaystyle \frac{1}{\sqrt{3}}}\end{array}\right)\text{and}D=\left(\begin{array}{rrr}-2& 0& 0\\ 0& -2& 0\\ 0& 0& 4\end{array}\right)$

4. ${\text{T}}_z$ is normal for all $z\in C,{\text{T}}_z$ is self-adjoint if and only if $z\in R$, and ${\text{T}}_z$ is unitary if and only if $|z|=1$ .

5. Only the pair of matrices in (d) are unitarily equivalent.

25. $2(\psi -\varphi )$

26.

1. (a) $\psi -{\displaystyle \frac{\varphi }{2}}$

2. (b) $\psi +{\displaystyle \frac{\varphi }{2}}$

27.

1. (a) $x={\displaystyle \frac{1}{\sqrt{2}}}{x}^{\prime }+{\displaystyle \frac{1}{\sqrt{2}}}{y}^{\prime }$ and $y={\displaystyle \frac{1}{\sqrt{2}}}{x}^{\prime }-{\displaystyle \frac{1}{\sqrt{2}}}{y}^{\prime }$

   The new quadratic form is $3{(x^{\prime })}^2-{(y^{\prime })}^2.$

2. (c) $x={\displaystyle \frac{3}{\sqrt{13}}}{x}^{\prime }+{\displaystyle \frac{2}{\sqrt{13}}}{y}^{\prime }$ and $y={\displaystyle \frac{-2}{\sqrt{13}}}{x}^{\prime }+{\displaystyle \frac{2}{\sqrt{13}}}{y}^{\prime }$

   The new quadratic form is $5{(x^{\prime })}^2-8{(y^{\prime })}^2.$

29.

1. (c) $Q=\left(\begin{array}{rrr}{\displaystyle \frac{1}{\sqrt{2}}}& {\displaystyle \frac{1}{\sqrt{3}}}& -{\displaystyle \frac{\sqrt{6}}{6}}\\ {\displaystyle \frac{1}{\sqrt{2}}}& -{\displaystyle \frac{1}{\sqrt{3}}}& {\displaystyle \frac{\sqrt{6}}{6}}\\ 0& {\displaystyle \frac{1}{\sqrt{3}}}& {\displaystyle \frac{\sqrt{6}}{3}}\end{array}\right)\text{and}R=\left(\begin{array}{ccc}\sqrt{2}& \sqrt{2}& 2\sqrt{2}\\ 0& \sqrt{3}& {\displaystyle \frac{\sqrt{3}}{3}}\\ 0& 0& {\displaystyle \frac{\sqrt{6}}{3}}\end{array}\right)$

2. (e) $x_1=3,x_2=-5,x_3=4$