A diagonal matrix can be expressed as a difference of two squares ($A=B^2 − C^2$ where $BC = CB = 0$) I got one interesting question from matrix theory. I tried but I am not finding any clue to solve this question.  I need help and suggestions. Thanks in advance.
Let $A$ be a real $n\times n$ matrix. We say that $A$ is a difference of two squares if there exist real $n\times n$ matrices $B$ and $C$ with $BC = CB = 0$ and $A = B^2 − C^2$.
Now If $A$ is a diagonal matrix, then I have to show that that it is a difference of two squares.

Perhaps this may help you .
Let $a$ be a real number. if $a\geq 0$ we can write $a = b^2 - c^2$ with $c = 0$ and if $a < 0$, we can write $a = b^2 − c^2$ with $b = 0$. Thus, in general, $a = b^2 − c^2$ with $b$ and $c$ real and $bc = 0$. Now if $A = \operatorname{diag}(a_1, a_2, . . . , a_n)$ is a real diagonal matrix, write each $a_i = b_i^{2}− c_i^2$ as above, with $b_i c_i = 0$, and define $B = \operatorname{diag}(b_1, b_2, . . . , b_n)$ and $C = \operatorname{diag}(c_1, c_2, . . . , c_n)$. Then, by the way diagonal matrices multiply, we conclude that $A = B^2−C^2$ and also that
$BC = CB = 0$.
