Note que se \(A\) é diagonalizável, então
\(A=MDM^{-1}=MD^{1/2}D^{1/2}M^{-1}=MD^{1/2}M^{-1}MD^{1/2}M^{-1}=\sqrt{A}\sqrt{A}\).
Logo, \(\sqrt{A}= MD^{1/2}M^{-1}\).
O cálculo dos autovalores retorna \(\lambda =2\), \(\lambda =2\) e \(\lambda=1\).
Os autovetores associados a \(\lambda =2\) são \((1,0,-1)^\prime\) e \((0,1,0)^\prime\).
O autovetor associado a \(\lambda =1\) é \((0,1,-1)^\prime\)
Logo, \(M=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 1\\
-1 & 0 & -1\\
\end{array}\right]\)
e
\(M^{-1}=\left[\begin{array}{ccc}
1 & 0 & 0\\
-1 & 1 & 1\\
1 & 0 & -1\\
\end{array}\right]\)
\(D=M^{-1}AM=\left[\begin{array}{ccc}
2 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1\\
\end{array}\right]\).
Logo,
\(\sqrt{D}=\left[\begin{array}{ccc}
\sqrt{2} & 0 & 0\\
0 & \sqrt{2} & 0\\
0 & 0 & 1\\
\end{array}\right]\)
Chegando a
\(\sqrt{A}=\left[\begin{array}{ccc}
\sqrt{2} & 0 & 0\\
\sqrt{2}-1 & \sqrt{2} & \sqrt{2}-1\\
1-\sqrt{2} & 0 & 1\\
\end{array}\right]\)