Similarity in two 2x2 Matrices and finding the S in A=SBS-1 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!
 A: $$S=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;\;\text{and}\;\;\;A=SBS^{-1}\iff AS=SB\iff$$
$$AS=\begin{pmatrix}1&1\\\!\!-2&5\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\;\;\,a+c&\;\;\,b+d\\-2a+5c&-2b+5d\end{pmatrix}=\begin{pmatrix}4a+b&a+2b\\4c+d&c+2d\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}4&1\\1&2\end{pmatrix}$$
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, 
$$S=\begin{pmatrix}1&\!\!-1\\2&0\end{pmatrix}\implies S^{-1}=\begin{pmatrix}0&\frac12\\\!\!-1&\frac12\end{pmatrix}$$
A: $A=PDP^{−1}=P(Q^{−1}BQ)P⟹S=PQ^{−1}$
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}
