Problem related with the similarity of matrices

Question

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.

Answer 1

HINT:

The matrix $A$ is similar only to itself. Thus $PAP^{-1}=A $ for all invertible $P.$
Use as $P$ the matrices: $$P=\begin{pmatrix}1 &1 \\ 0 &1 \end{pmatrix} \ \ \text{and} \ \ P=\begin{pmatrix}1 &0 \\ 1 &1 \end{pmatrix}.$$

Answer 2

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 $x$.

Answer 3

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).