Sums of Nilpotent Matrices

Question

Let $A =$ diag$(a_1,a_2,…,a_n)$, where the sum of all $a_i$’s is zero.

Show that A is a sum of nilpotent matrices.

My idea:

$\begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$ + $\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$ Where $\begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}^2$ = $\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}^2$ = $0_2$

Extending the 2 $\times$ 2 matrix, we get

$\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & -2 \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 & 0 & 0 \\ -1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1 \end{bmatrix}$ + $\begin{bmatrix} 1 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix}$

Where $\begin{bmatrix} 1 & 1 & 0 & 0 \\ -1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1 \end{bmatrix}^2$ = $\begin{bmatrix} 1 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix}^2$ = $0_4$

Kindly correct me if I’m wrong, and also, please help me how to generalized this. Thank you in advance.

Answer 1

The diagonal matrices of the form $\operatorname{diag}(0,\ldots,0,1,0,\ldots,0,-1)$ form a basis of the space of the matrices with null trace. In fact, a matrix with null trace is $\operatorname{diag}(a_1,\ldots,a_n)$ with $a_1+a_2+\cdots+a_n=0$, which is equal to$$a_1\operatorname{diag}(1,0,0,\ldots,0,-1)+a_2\operatorname{diag}(0,1,0,\ldots,0,-1)+\cdots+a_{n-1}\operatorname{diag}(0,0,\ldots,0,1,-1).$$Now, you can apply your idea to each matrix of the type $\operatorname{diag}(0,\ldots,0,1,0,\ldots,0,-1)$.

Answer 2

My answer follows @JoséCarlosSantos' post.
It is enough to notice that for all $1 \leq i < j \leq n$, $$E_{i, i} -E_{j, j} = \left(E_{i, j}\right) + \left(- E_{j, i}\right) + \left(E_{i, i} -E_{i, j} + E_{j, i} - E_{j, j}\right)$$ and $$\left(E_{i, i} -E_{i, j} + E_{j, i} - E_{j, j}\right)^{2} = 0$$