Nilpotent matrix and similar matrix

Question

If $A$ is a complex matrix of order $n$ and $A$ is nilpotent, that is, there exists a positive integer $s$ such that $A^s=0$. Let's say that $e^A = \sum_{k=0}^{\infty} \frac{A^k}{k!}$. Prove that $e^A$ is similar to $I_n+A$, where $I_n$ is the identity matrix of order $n$.

Answer 1

This is equivalent to show that $A$ is similar to $e^A - I = A + \frac{A}{2!} + \cdots$. We show a more general statement that $A$ is similar to $B = A + a_2A^2 + a_3A^3 + \cdots$ for arbitrary coefficients $a_2, a_3, \cdots$.

It's enough to show this when $A$ has a single Jordan block, i.e. there is a vector $v_0$ such that $\{ v_0, v_1=Av_0, v_2=Av_1, \cdots \}$ form a basis of $\mathbb C^n$. Now it's enough to show that $v_0, Bv_0, B^2v_0, B^3 v_0$ are linearly independent. Note that the coefficients of $v_i$ in $B^jv_0$ is $\delta_{ij}$ for $i\le j$. The matrix of $v_0, Bv_0, \cdots$ under the basis $v_0, v_1, \cdots$ is triangular with only $1$ on the diagonal, hence they are linearly independent.

Answer 2

It suffices to show that this holds in the case that $A$ is a nilpotent Jordan block (of size $n$) with eigenvalue $0$.

Note that it is equivalent to show that $e^A - I$ is similar to $A$, and that that a matrix $B \in \Bbb C^{n \times n}$ is similar to a nilpotent Jordan block (of size $n$) if and only if it is nilpotent with $B^{n-1} \neq 0$. So, let $B = e^A - I$. By considering the series expansion $B = A\sum_{k=0}^\infty \frac{A^k}{(k+1)!}$, note that for all exponents $m$, we have $$ B^m = A^m (I + a_1 A^1 + \cdots + a_{n-1}A^{n-1}) $$ for some coefficients $a_1,\dots,a_{n-1} \in \Bbb C$. Conclude therefore that $B^{n-1} = A^{n-1} \neq 0$ and that $B^{n} = 0$. Thus, $B$ must be similar to a Jordan block of size $n$, which is to say that it is similar to $A$.