Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over the matrix. Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over $\mathbb{C}$ to \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix} 
 A: Elementarily, assume $$N = \begin{bmatrix}a & b\\c & d\end{bmatrix} $$
Then
$$0=N^2 = \begin{bmatrix}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix} $$
Now if $b=0$, you get 
$$0=N^2 = \begin{bmatrix}a^2 & 0\\c(a+d) & d^2\end{bmatrix} $$
so that $a=d=0$, and the result follows.
Otherwise, you must conclude $a=-d$ so 
$$0=N^2 = \begin{bmatrix}a^2+bc & 0\\0 & a^2+bc\end{bmatrix} $$
giving that $c=-\frac{a^2}{b}$ so the original matrix is
$$N = \begin{bmatrix}a & b\\-\frac{a^2}{b} & -a\end{bmatrix} $$
and then observe that
$$\begin{bmatrix}0 & 1\\\frac{1}{b} & -\frac{a}{b}\end{bmatrix}
  \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}
  \begin{bmatrix}a & b\\1 & 0\end{bmatrix} = N$$
A: Use the Jordan normalform and write your matrix as $D+M$ where $D$ is a diagonal matrix and $M$ is a nilpotent matrix. So there are only 2 cases to check.
We have $$N^2=(D+M)^2 = D^2 + DM+ MD+ M^2$$
As $DM=MD$ and $M^2=0$ as $M$ is a nilpotet $2 \times 2$ matrix we have:
$$N^2=D^2+ 2 MD$$
As $MD$ will have only zeroes on the diagonal we know that $D=0$ (else $D^2 \neq 0$). 
As we looked at the Jordan Normalform we need to check two cases. The first one is 
$$M=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$$ 
the second is 
$$M=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$$
A: Let $J=\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}$. When $N\not=0$, there exists a nonzero vector $x$ such that $Nx\not=0$. Clearly $x$ and $Nx$ are linearly independent, otherwise $Nx=cx$ for some $c\neq0$ and in turn $0=N^2x=N(Nx)=cNx\neq0$, which is a contradiction. Therefore $P=(x,Nx)$ is invertible. Now you may verify that $NP=(Nx,N^2x)=(Nx,0)=PJ$. So, $N=PJP^{-1}$.
