Complex matrix decomposition into the sum of a diagonalizable and a nilpotent matrix. Let $A$ be any $n \times n$ complex matrix. Prove that $A$ can be written as $A = B + N$
where $B$ is diagonalizable, $N$ is nilpotent (some power is the zero matrix) and the
matrices $B$ and $N$ commute.
 A: Every matrix can be decomposed into Jordan form .
So we have $A=VJV^{-1}$. 
Jordan matrix is a sum of some diagonal matrix $D$ and some upper triangular matrix $N$ with $0$'s on diagonal. Because on diagonal there are only zeros all eigenvalues are also $0$.
Hence $N$ is nilpotent (easy proof from CH theorem) - we see that
any triangular matrix with 0's along the main diagonal is nilpotent.
We have   
$A=V(D+N)V^{-1}$  
and because similarity preserves eigenvalues so $N_v=V  N V^{-1}$ is also nilpotent.
You can also check that $(V N V^{-1})^k=  V N^k V^{-1} $ hence for some $k$ it holds ${N_v}^k=0$.
Commutation. 
To $VDV^{-1}$ commute with $VNV^{-1}$ it's enough to show that $D$ commute with $N$ i.e. $ND=DN$. This commutation is assured by the construction of Jordan block. 
Take for example $D_i= \begin{bmatrix}  \lambda_i & 0 & 0 \\ 0 & \lambda_i& 0 \\ 0 & 0 & \lambda{_i}  \end{bmatrix}$ and $N_i= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0  & 1\\ 0 & 0 & 0  \end{bmatrix}$.
and you will see that $D_iN_i=N_iD_i$.     
Multiplication is commutative between these parts of Jordan block because $D_i=\lambda_i I$,  consequently multiplication of $N_i$ is just a multiplication of $N_i$ by a scalar $\lambda_i$.
A: 
Every $n\times n$ matrix $A$ is (unitarily) similar to a triangular matrix $T$.

If the above statement is accepted, then the requested decomposition is easy to write down: the triangular matrix $T$ can be seen as $T_0+N_0$, where $T_0$ is diagonal with the same entries on the diagonal as $T$. Then $N_0=T-T_0$ is nilpotent. Write $A=STS^{-1}$ and finish up.
Proof of the main statement.
Choose a norm $1$ eigenvector for $A$, call it $v_1$ with $Av_1=\lambda v_1$, and complete it to an orthonormal basis $\{v_1,\dots,v_n\}$ of $\mathbb{C}^n$. If $U_0=[v_1\ \dots\ v_n]$, then $U_0$ is unitary and
$$
U_0^HAU_0
$$
has its first column in the form
$$
\begin{bmatrix} \lambda \\ 0 \\ \vdots \\ 0\end{bmatrix}
$$
Remove the first row and column: by inductive hypothesis, the $(n-1)\times(n-1)$ matrix $A_1$ that remains is unitarily similar to a triangular matrix, say $T_1=U_1^HA_1U_1$ is triangular.
Then
$$
\begin{bmatrix}
\lambda & \dots \\
0 & T
\end{bmatrix}=
\begin{bmatrix} 1 & 0 \\ 0 & U_1 \end{bmatrix}^H
U_0^H A U_0
\begin{bmatrix} 1 & 0 \\ 0 & U_1 \end{bmatrix}
$$
Thus we can write $T=U^HAU$, with $U$ unitary and $T$ triangular.
Note: $X^H$ stands for the Hermitian transpose of the matrix $X$.
