Suppose $N\in M_{3\times 3}^{\mathbb{C}}$ is a nilpotent matrix. Prove that there exists $A\in M_{3\times 3}^{\mathbb{C}}$ such that $A^2=I+N$. Hint: find $A$ in the form $A=P(N)$ where $P$ is a polynomial in $\mathbb{C}[x]$.
I really can't think of a way to prove this. I know that $N^3=0$, so I tried combining this fact with $A^2=I+N$. I deduced that $(A^2-I)^3=0$, and thus the minimal polynomial of $A$ must include the terms $(x-1)(x+1)$. Any suggestions?
We can try approximations that turn out to be exact: As a power series (Taylor), we have $\sqrt{1+x}=1+\frac 12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4\pm\ldots$ so if we know that $x^3=0$ we get that $\sqrt{1+x}=1+\frac12x-\frac18x^2$. So letting $$A=I+\frac12N-\frac18N^2$$ wins.
If you follow the hint, you can find the expansion "algebraically" without turning to Taylor series. Write $A = a_0I + a_1N + a_2N^2$ (no need to consider higher powers since $N$ is a $3 \times 3$ nilpotent matrix). Then (again, ignoring powers larger than $2$) we want to find $a_i$ such that
$$ A^2 = a_0^2 \cdot I + 2a_0a_1 \cdot N + (2a_0a_2 + a_1^2) \cdot N^2 = I + N.$$
By comparing coefficients, you can easily find two options for $A$.
