Let $A$ be a $2×2$ non zero matrix with entries in$\Bbb C$ such that $A^2 = 0$. Which of the following statements must be true ?
$PAP^{-1}$ is diagonal for some invertible $2×2$ matrix $P$ with entries in $\Bbb R$
$A$ has two distinct eigen values in $\Bbb C$
$A$ has only one eigen value in $\Bbb C$ with multiplicity $2$
$Av = v$ for some $v \in \Bbb C^2, v \neq 0$
My Attempt: Option 1 If I take $P = I$ then $PAP^{-1} = IAI^{-1} = A$ which is non zero nilpotent matrix and it is not diagonalizable . So $PAP^{-1}$ is not diagonalizable Hence option 1 is not true.
Option 2,3: $A$ is nilpotent matrix. So option 3 is true and 2 is false.
Option 4:
Since $A$ is non zero nilpotent matrix. So it is of the form $$A = \begin{pmatrix} 0&a_{12}\\0&0\end{pmatrix}$$ or $$A = \begin{pmatrix} 0&0\\a_{21}&0\end{pmatrix}$$ Take $v = (\alpha , \beta)^T$ with $v \in \Bbb C^2$ and $\alpha,\beta \neq 0$ and $v \neq 0$ Now $Av = v$ implies $\begin{pmatrix} 0&a_{12}\\0&0\end{pmatrix}\begin{pmatrix}\alpha \\ \beta \end{pmatrix} =\begin{pmatrix}\alpha \\ \beta \end{pmatrix} \implies \begin{pmatrix}a_{12}\beta \\ 0\end{pmatrix} = \begin{pmatrix}\alpha \\ \beta \end{pmatrix} \implies \alpha = \beta = 0$ Which is a contradiction as we assumed $\alpha, \beta \neq 0$. So option 4 is false. Please correct me if I'm on wrong track or provide alternate solution if exist.