Exercise 1.6.23

The $X=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill \end{array}\right]$, so $X^{-1}=\left[\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill -\frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \end{array}\right]$. And $\Lambda =\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 9\hfill \end{array}\right]$.

We have $A=\mathit{X\Lambda }{X}^{-1}$ and square root of $A$ is: $R=X\sqrt{\Lambda }{X}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill -\frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$.

Ther is no real matrix square root of $B$ because the first eigenvalue is $-1<0$, so no real square root here.