Answer
See below
Work Step by Step
a) Since $D=diag(\lambda_1,\lambda_2,...\lambda_m)$
then $ \sqrt D=diag(\sqrt \lambda_1,\sqrt \lambda_2,...\sqrt \lambda_m)$
We obtain: $\sqrt D.\sqrt D\\
=diag(\sqrt \lambda_1,...\sqrt \lambda_m).diag(\sqrt \lambda_1,...\sqrt \lambda_m)\\
=diag(\lambda_1,...\lambda_m)$
b) $(S\sqrt DS^{-1})^2\\
=(S\sqrt DS^{-1})(S\sqrt DS^{-1})\\
=S\sqrt D\sqrt DS^{-1}\\
=SDS^{-1}\\
=A$
$S\sqrt DS^{-1}$ is a square root of A.
c) The given system can be written as $x′ = Ax$,
where $A=\begin{bmatrix}
6 & -2\\
-3 & 7
\end{bmatrix}$
The transformed system is $y′ = (S−1AS)y$ where $x = Sy$
To determine S, we need the eigenvalues and eigenvectors of A. The characteristic polynomial of A is
$p(\lambda)=\begin{bmatrix}
6-\lambda & -2\\
-3 & 7-\lambda
\end{bmatrix}=\lambda^2-13\lambda+36=(\lambda-4)(\lambda-9)$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
$\lambda_1=4: v=r(1,1) \\
\lambda_2=9: v=s(2,-3)\\$
Set $S = \begin{bmatrix}1 &2\\
2 & -3
\end{bmatrix}$
then from Theorem 7.3.4, $S−1AS = diag(4,9)$
so that the system becomes:
$\sqrt D=S\begin{bmatrix}
2 & 0\\
0 & 3
\end{bmatrix}$
so $S^{-1}=\begin{bmatrix}
\frac{3}{5} & \frac{2}{5}\\
\frac{1}{5} & -\frac{1}{5}
\end{bmatrix}$
Hence $\sqrt A=S\sqrt DS^{-1}=\begin{bmatrix}
\frac{12}{5} & \frac{-2}{5}\\
\frac{-3}{5} & \frac{13}{5}
\end{bmatrix}$
