Logarithm of $I_n-N$ for $N$ nilpotent

Question

Prove that $$L=-\sum_{k=1}^{n-1} \frac{N^{k}}{k}$$ is a logarithm of the matrix $I_n-N\!$, where $I_n$ denotes the identity, and $N$ is nilpotent such that $N^n=0$.

More precisely, I want to show that $\exp(L)=I_n-N$. I have not seen the logarithmic function for matrices yet.

This is the last step that has been used in this argument, but I haven't been able to prove it. Any hints?

Answer 1

Let $L(t) = - \sum_{k \geq 1} \frac{t^k N^k}{k}$. There are no convergence issues for $L(t)$ because $N$ is nilpotent. Now define $f(t) = \exp(L(t))$, and note that $f$ satisfies the differential equation $$ f'(t) = L'(t) \exp(L(t)) = \left(-N\sum_{k \geq 0} t^k N^k\right) \exp(L(t)) = -N(I - tN)^{-1} f(t).$$ However, the function $g(t) = I - tN$ also satifies this differential equation: $$ g'(t) = -N = -N(I-tN)^{-1}g(t),$$ with the same initial conditions $g(0) = f(0) = I$, and hence they must be equal. The wanted equality is case $t=1$.

Answer 2

Here's a hint to a possible approach, although it's probably nowhere near the cleanest:

You know that the exponential series converges for all matrices. Consider

$$L^m=(-1)^m\left(\sum_{k=1}^{n-1} \frac{N^k}k\right)^m.$$

You have

$$\left(\sum_{k=1}^{n-1} \frac{N^k}{k}\right)^m=\sum_{\mathbf k\in [n-1]^m}\frac{N^{\sum_{i=1}^m k_i}}{\prod_{i=1}^m k_i},$$

where $\mathbf k$ is a vector of $m$ values of $k$ and $[r]$ denotes the set of integers from $1$ to $r$. You can rewrite this, using $N^n=0$, as

$$\sum_{j=0}^{n-1}N^j\sum_{\substack{\mathbf k\in [n-1]^m\\ \mathbf k\cdot\mathbf 1=j}}\frac1{\prod_{i=1}^m k_i}.$$

Write $$a_{j,m}=\sum_{\substack{\mathbf k\in [n-1]^m\\ \mathbf k\cdot\mathbf 1=j}}\frac1{\prod_{i=1}^m k_i}.$$

Since the exponential series is always convergent, we have \begin{align*} \exp(L) &=\sum_{m=0}^\infty \frac{L^m}{m!}\\ &=\sum_{m=0}^\infty (-1)^m\frac{1}{m!}\sum_{j=0}^{n-1} N^j a_{j,m}\\ &=\sum_{j=0}^{n-1} N^j\sum_{m=0}^\infty \frac{(-1)^ma_{j,m}}{m!}. \end{align*}

So, we need to show that $$\sum_{m=0}^\infty \frac{(-1)^ma_{j,m}}{m!}$$ is $1$ at $j=0$, $-1$ at $j=1$, and zero for larger $j$. See if you can prove this using generating functions and looking at this sum in terms of partitions of $j$.

Answer 3

Here's a somewhat sneaky proof. I'll assume we're working over $\mathbb{C}$ and already know facts about the power series for $\exp$ and $\log$ of real numbers.

Let $R = \mathbb{C}[\![z]\!]$ be the ring of formal power power series over $\mathbb{C}$. Let $e := \sum_{i=0}^\infty z^i / i!$ and $l: = -\sum_{i=1}^\infty z^i/i$. Since the constant term of $l$ is zero, the formal composition $q := e \circ l$ is a well-defined element of $R$.

Claim $q = 1-z$.

Proof. Consider $e$ and $l$ as power series defining holomorphic functions on open subsets of $\mathbb{C}$. Since $e$ converges absolutely everywhere and $l$ converges absolutely on a small open disk around $0$, $q$ converges on a small open disk around $0$. By real analysis, $q(x) = 1-x$ for all real numbers $x$ in this disk. By the uniqueness theorem of complex analysis, $q = 1-z$.

Now consider what happens to the identity $e \circ l = 1-z$ when we mod out by the ideal $(z^n) \subseteq R$. The quotient ring is $R/(z^n) \cong \mathbb{C}[z]/(z^n)$, and the composition of polynomials in this quotient is given by the same formal rule as in $R$. Thus, the identity $e(l(z)) = 1-z$ holds in $\mathbb{C}[z]/(z^n)$. But $\mathbb{C}[z]/(z^n)$ is the universal $\mathbb{C}$-algebra containing a nilpotent element of degree $n$, so this formal identity must hold for all nilpotent elements of all $\mathbb{C}$-algebras! In particular, it holds for your matrix $N \in M_r(\mathbb{C})$.