Question 7.5.8: For A =[3 -5 1 -1], find an invertible 2 × 2 matrix S such th......

We will use the method outlined in Theorem 7.5.3:

f_A(λ)=λ^2-2λ+2,  so that  λ_{1,2}=\frac{2±\sqrt{4-8}}{2}=1±i.

Now

E_{1+i}=ker\begin{bmatrix}2-i&-5\\1&-2-i\end{bmatrix}=span\begin{bmatrix}-5\\-2+i\end{bmatrix},

and

\begin{bmatrix}-5\\-2+i\end{bmatrix}=\begin{bmatrix}-5\\-2\end{bmatrix}+i\begin{bmatrix}0\\1\end{bmatrix},  so that  \vec{w}=\begin{bmatrix}0\\1\end{bmatrix},    \vec{v}=\begin{bmatrix}-5\\-2\end{bmatrix}​.

Therefore,

S^{-1}AS=\begin{bmatrix}1&-1\\1&1\end{bmatrix},  where    S=\begin{bmatrix}0&-5\\1&-2\end{bmatrix}.