Problem related with the similarity of matrices I came across this problem which says:
Let $A$ be a $2 \times 2$ matrix such that only $A$ is similar to itself.Then  show that A is a scalar matrix, that is $A$ is of the form \begin{pmatrix}
a &0 \\ 
0 & a
\end{pmatrix}
?
My attempts: 
 Since $A$ is similar to itself,there exists an invertible matrix P such that A=
$P^{-1}AP$. Then I tried to solve it by choosing A and P of the form \begin{pmatrix}
a &b \\ 
c & d
\end{pmatrix}
and \begin{pmatrix}
x &y \\ 
z & w
\end{pmatrix}
respectively. But I could not get the desired result. Please help. Thanks everyone in advance for your time.
 A: HINT:
The matrix $A$ is similar only to itself. Thus $PAP^{-1}=A $ for all invertible $P.$ 
Use as $P$ the matrices:
$$P=\begin{pmatrix}1 &1 \\ 0 &1 \end{pmatrix} \ \ \text{and} \ \ P=\begin{pmatrix}1 &0 \\ 1 &1 \end{pmatrix}.$$
A: Alternatively, that $A$ is only similar to itself means $PA=AP$ for all invertible matrix $P$. Therefore
$$
\begin{pmatrix}a+cx&b+dx\\ cy&dy\end{pmatrix}=
\underbrace{\begin{pmatrix}1&x\\0&y\end{pmatrix}}_{P}
\underbrace{\begin{pmatrix}a&b\\c&d\end{pmatrix}}_{A}=
\begin{pmatrix}a&b\\c&d\end{pmatrix}
\begin{pmatrix}1&x\\0&y\end{pmatrix}=
\begin{pmatrix}a&ax+by\\c&cx+dy\end{pmatrix}
$$
for all $y\not=0$ and all $x$.
A: Do you know that every $2\times2$ matrix is similar either to a matrix of the form $$A=\pmatrix{a&0\cr0&b\cr}$$ or to a matrix of the form $$B=\pmatrix{a&1\cr0&a\cr}?$$ If so, then all you have to show is that $A$ is similar to $$\pmatrix{b&0\cr0&a\cr}$$ and $B$ is similar to $$\pmatrix{a&0\cr1&a\cr}$$ (and the latter follows from the theorem that every matrix is similar to its transpose). 
