Similarity in two 2x2 Matrices and finding the S in A=SBS-1

Question

I am doing something wrong here and I am not sure what.

The object of the exercise is to find the S for similar matrices $A$ and $B$.

$A=SBS^{-1}$

with $B=\begin{pmatrix}4& 1\\1& 2\end{pmatrix}$

and $A=\begin{pmatrix}1& 1\\-2& 5\end{pmatrix}$

I assumed it had something to do with the Eigenvectors and a basis built from them (because that is the current topic in the course), but it isn't working out. I only need the $S$, but I'd like the $S^{-1}$ as well, to better understand what's going on.

I've calculated the diagonal matrices for both and the corresponding $T$ and $T^{-1}$ transformation matrices, but no combination of them multiplied with $B=A$. Am I over complicating things?

Thanks in advance!

Answer 1

$$S=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;\;\text{and}\;\;\;A=SBS^{-1}\iff AS=SB\iff$$

$$AS=\begin{pmatrix}1&1\\\!\!-2&5\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\;\;\,a+c&\;\;\,b+d\\-2a+5c&-2b+5d\end{pmatrix}=\begin{pmatrix}4a+b&a+2b\\4c+d&c+2d\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}4&1\\1&2\end{pmatrix}$$

Well, now you "just have" a linear system in $\;4\;$ unknowns:

$$\begin{cases}a+c=4a+b\\b+d=a+2b\\-2a+5c=4c+d\\-2b+5d=c+2d\end{cases}$$

Solving it you get, for example,

$$S=\begin{pmatrix}1&\!\!-1\\2&0\end{pmatrix}\implies S^{-1}=\begin{pmatrix}0&\frac12\\\!\!-1&\frac12\end{pmatrix}$$

Answer 2

$A=PDP^{−1}=P(Q^{−1}BQ)P⟹S=PQ^{−1}$
with $P$ and $Q$ built from the Eigenvectors.

$P$ \begin{matrix} 1-\sqrt{2}& 1+\sqrt{2}\\ 1& 1& \end{matrix}

$P^{-1}$ \begin{matrix} -1/2\sqrt{2}& (1+\sqrt{2})/2\sqrt{2}\\ 1/2\sqrt{2}& (-1+\sqrt{2})/2\sqrt{2}& \end{matrix}

$Q$ \begin{matrix} 1+(1/\sqrt{2})& 1-(1/\sqrt{2})\\ 1& 1& \end{matrix}

$Q^{-1}$ \begin{matrix} 1/\sqrt{2}& (1/2)-(1/\sqrt{2})\\ -1/\sqrt{2}& (1/2)+(1/\sqrt{2})& \end{matrix}