$A$ 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?
Calculate the Taylor series expansion of $\sqrt{1-t}$ up to the $t^3$ term. You will get $B$ by substituting in $t=A$.
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 $B^2=I-A$.
In matrices, $XY=0$ doesn't imply $X=0\lor Y=0$, nor does $X^4=0$ imply $X=0$. If we had $B=\pm I$, we'd have $A=I-B^2=0$, which isn't the most general solution of $A^4=0$. @tkf's answer give $B=I-A/2-A^2/8-A^3/16$ as one solution (but we could multiply that by $-1$ to get another). You can then verify $1-B^2=A+O(A^4)=A$.
