<s>[INST] Sums of Nilpotent Matrices 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.
 [/INST] 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 $\operatorname{diag}(0,\ldots,0,1,0,\ldots,0,-1)$.
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