TITLE: If $N$ is nilpotent then there exists $A$ such that $A^2=I+N$
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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 $A$.