Let $A$ be a $2\times 2$ complex matrix,then show that $A$ is similar over $\mathbb C$ to either
$\begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix}$ or $\begin{bmatrix} a & 0 \\ 1 & a \\ \end{bmatrix}$.
Proof: If $A$ is diagonalizable,then it is similar to the first form.
If $A$ is not diagonalizable then $A$ has only one eigenvalue with $1$-dimensional eigenspace.(Since,every complex matrix has an eigenvalue).
Let,$\mathcal B=\{a_1,a_2\}$ be a basis such that $a_2$ is an eigenvector of $T$.($T$ is the linear trasnformation such that $T$ with respect to standard basis $\{(1,0),(0,1)\}$ is $A$).
Then,$T(a_1)=c_1a_1+c_2a_2$ and $T(a_2)=\lambda a_2$ where $\lambda$ is the only eigenvalue of $T$.
Then $[T]_{\mathcal B}=\begin{bmatrix} c_1 & 0 \\ c_2 & \lambda \\ \end{bmatrix}$
But, in triangular matrix,diagonals are eigenvalues,so $c_1=\lambda$.
So,$[T]_{\mathcal B}=\begin{bmatrix} \lambda & 0 \\ c_2 & \lambda \\ \end{bmatrix}$
Then,we change our basis to $\mathcal B'=\{a_1,c_2a_2\}$ ,notice $c_2\neq 0$ as $T$ is not diagonalizable under this case and if $c_2=0$ , then $[T]_{\mathcal B}$ is a diagonal matrix with diagonals $\lambda$.
Then we have $[T]_{\mathcal B'}=\begin{bmatrix} \lambda & 0 \\ 1 & \lambda \\ \end{bmatrix}$
Is my solution alright or it could be done in a more simple way.
Note: I do not want to use $T=D+N$ decomposition in $\mathbb C$.I only want solution using basic knowledge of eigenvalues.