$A^2 = I+N$ when $N$ is Nilpotent over $C$

Question

Prove that an $A_{3×3}$ matrix exists over $\mathbb{C}$ that

$A^2 = I+N$ And $N$ is nilpotent ($3×3$ over $\mathbb{C}$)

the hint says: find $A$ as $P(N)$ when $P$ is a polynomial over $\mathbb{C}$

Answer 1

If you have to find $A$ for given $N$, you can do as follows:

First, we should have $N^3=O$ since $N$ is nilpotent and its minimal polynomial, which has degree less or equal to 3, should divide $x^k$ for some $k$. Now try $A=aI+bN+cN^2$ and $A^2=a^{2}I+2abN+(b^{2}+2ac)N^{2}=I+N$. Can you find appropriate $a,b,c$?

Answer 2

We define $$A=I+\dfrac{1}{2}N-\dfrac{1}{2}\cdot\dfrac{1}{2}N^2+\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}N^3-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}N^4+\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}\cdot\dfrac{7}{2}N^5-\cdots$$which is the Taylor series expansion of $\sqrt{I+N}$. Also since $N$ is nilpotent we have $$N^k=0\quad,\quad\text{for some k}$$therefore $$N^n=0\quad,\quad l\ge k$$ and the terms of the above expansion would get zero after a while. So we have a finite-degree polynomial representation of $A$ respect to $N$.