I have been given as an exercise the following statement to prove, but I have no idea where to start and how to go about proving it.
If $N$ is a nilpotent matrix in $M_n(\mathbb{C})$ and $r$ is a natural number, prove that there exists $A$ in $M_n(\mathbb{C})$ s.t. \begin{align*} A^r = N + I_n. \end{align*}
Any help will be much appreciated.
(btw we have just been taught the primary decomposition theorem and the Jordan canonical form.)
On
Here is a quick solution using the binomial series. Define $A$ by
$$ A = \sum_{k=0}^{\infty} \binom{1/r}{k} N^k. \tag{1} $$
Here, $ \binom{a}{k} = \frac{a(a-1)\cdots(a-k+1)}{k!}$, where $a \in \mathbb{C}$ and $k \in \{0, 1, \ldots\}$, denotes the generalized binomial coefficient. Since $N$ is nilpotent, the above series is actually a finite sum and hence is well-defined. Moreover,
\begin{align*} \sum_{k=0}^{n} \binom{a}{i}\binom{b}{n-i} &= \binom{a+b}{n} \tag{2} \end{align*}
continues to hold for arbitrary $a, b \in \mathbb{C}$ and $n \in \{0, 1, \ldots\}$. Using $\text{(2)}$ repeatedly, we get
\begin{align*} A^r &= \sum_{l=0}^{\infty} \sum_{\substack{k_1+\cdots+k_r=l \\ k_1, \ldots, k_r \geq 0 }} \binom{1/r}{k_1}\cdots\binom{1/r}{k_r} N^l \\ &= \sum_{l=0}^{\infty} \binom{1}{l} N^l \\ &= I + N, \end{align*}
Here's an approach that uses Jordan canonical form. To begin, consider the case where $N \in M_n(\Bbb C)$ is a single Jordan block associated with $0$.
Denote $J_1 = J_1^{(n)} = \text{Id} + N$, which is the Jordan block of the same shape associated with $1$. Note that $J_1^r$ is similar to $J_1$. To see that this is the case, compute $$ J_1^n = \pmatrix{1 & \binom{n}1 & \binom n2 & \cdots\\ 0 & 1 & \binom n1 & \vdots\\ \vdots & \ddots & \ddots & \binom n1\\ 0 & \cdots & 0 & 1 } $$ where $\binom nk$ denotes a binomial coefficent. Note that $J_1^r$ has $1$ as its only eigenvalue and that the rank of $J_1^r - \text{Id}$ is equal to $n-1$. Conclude that the only possible Jordan form for $J_1^r$ is $J_1$.
Thus, there exists an invertible matrix $P$ such that $PJ_1^rP^{-1} = J_1$. However, this means that $(PJ_1P^{-1})^r = J_1$. Thus, the matrix $A_0^{(n)} := PJ_1P^{-1}$ satisfies $[A_0^{(n)}]^r = J_1^{(n)}$.
Now, consider an arbitrary nilpotent $N$. Note that $\text{Id} + N$ has 1 as its only eigenvalue, so its Jordan form can be written as $$ Q^{-1}(\text{Id} + N)Q = \pmatrix{J_1^{(n_1)}\\ & \ddots\\ && J_1^{(n_k)}} $$ for suitable $n_1,\dots,n_k$ with $n_1 + \cdots + n_k = n$. This Jordan-form matrix has an $r$-th root: denote $$ \tilde A = \pmatrix{A_0^{(n_1)}\\ & \ddots \\ && A_0^{(n_k)}}. $$ Verify that $\tilde A^r = Q^{-1}(\text{Id} + N)Q$. Thus, we have $$ [Q \tilde A Q^{-1}]^r = \text{Id} + N. $$ Thus, we have proved the desired result with $A = Q \tilde A Q^{-1}$.
An interesting alternative approach via the exponential map:
We can show that $\exp$ defines a bijection between the nilpotent and unipotent matrices. In particular, it suffices to consider the power series definitions of $\exp$ and $\ln$ in order to show that the functions $N \mapsto \exp(N) - I_n$ and $N \mapsto \ln(I_n + N)$ are inverse functions over the set of nilpotent matrices, noting that all power series over $N$ are finite by the nilpotency of $N$.
From there, we may find a nilpotent matrix $M$ such that $\exp(M) = \text{Id} + N$. The properties of the matrix expoenential ensure that $A = \exp(M/r)$ satisfies $A^r = \exp(M) = \text{Id} + N$, which is what we wanted.
A possible approach that avoids Jordan canonical form:
Outline: Suppose that $N^{k-1} \neq 0$ and $N^k = 0$. Take $A = \text{Id} + a_1 N + a_2 N^2 + \cdots + a_{k-1}N^{k-1}$. Expand $A^r$ and argue that there exist $a_1,\dots,a_{k-1} \in \Bbb C$ such that $A^r = \text{Id} + N$.
The argument for $r = 2$ would go along the lines of what I've laid out here.