Exercise 1.8.20

$A=\left[\begin{array}{cc}\hfill -2\hfill & \hfill -6\hfill \\ \hfill 6\hfill & \hfill 2\hfill \end{array}\right]$, it has eigenvalues of ${\lambda }_1=4\sqrt{2}i,{\lambda }_2=-4\sqrt{2}i$, which are complex numbers. The eigenvectors are $x_1=\left[\begin{array}{c}\hfill \frac{2\sqrt{2}i-1}{3}\hfill \\ \hfill 1\hfill \end{array}\right]$and $x_2=\left[\begin{array}{c}\hfill \frac{-2\sqrt{2}i-1}{3}\hfill \\ \hfill 1\hfill \end{array}\right]$. We have $X=\left[\begin{array}{cc}\hfill \frac{2\sqrt{2}i-1}{3}\hfill & \hfill \frac{-2\sqrt{2}i-1}{3}\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

So $A^k=X{\Lambda }^k{X}^{-1}=X\left[\begin{array}{cc}\hfill {(4\sqrt{2}i)}^k\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {(-4\sqrt{2}i)}^k\hfill \end{array}\right]X^{-1}$.

So $A^k$ will have complex singular values and singular vectors.

$A^{T}A=\left[\begin{array}{cc}\hfill 40\hfill & \hfill 24\hfill \\ \hfill 24\hfill & \hfill 40\hfill \end{array}\right]$, this is a real symmetric matrix, it is also positive definite. So it has positive eigenvalues and real eigenvectors. Clearly, its singular values are real and positive, its singular vectors are real. This won’t satisfy (a) (b) with $A^k$ which has complex values.