<s>[INST] Find matrix $B$ such that $B^2 = I - A$ 
$A$ is a squared matrix such that $A^4 = \mathbf{0}$. Find $B$ in form of $A$ such that $B^2 = I - A$.

I tried in this way:
\begin{gather*}
A = I - B^2 \\
A ^ 4 = ( I - B)^4 (I + B)^4
\end{gather*}
So $I = B$ or $- I = B$
But the question wants $B$ as a form of $A$.
any help?
 [/INST] Assume $B$ can be written as a polynomial in $A$. That is,
\begin{align*}
B = c_0I+c_1A+c_2A^2+c_3A^3
\end{align*}
for unknown constants $c_0,c_1,c_2$, and $c_3$. Higher powers vanish since $A^4=0$. The relation $B^2=I-A$ gives
\begin{align*}
I-A = B^2 = c_0^2I+2c_0c_1A+(2c_0c_2+c_1^2)A^2+2(c_0c_3+c_1c_2)A^3.
\end{align*}
Comparing coefficients on either side yields $c_0=1$, $c_1=-\tfrac{1}{2}$, $c_2=-\tfrac{1}{8}$, and $c_3=-\tfrac{1}{16}$. So
\begin{align*}
B = I-\tfrac{1}{2}A-\tfrac{1}{8}A^2-\tfrac{1}{16}A^3
\end{align*}
solves $B^2=I-A$.
</s>