Is it true that nilpotent always has some eigenvalue?

Question

I understand that if a nilpotent matrix has some $\lambda$ eigenvector, then it implies that $\lambda=0$ because if $$Ax = \lambda x \\ A^2x = \lambda^2x \\ A^3x=\lambda^3x \\ \vdots \\ 0=A^k=\lambda^k\\ \lambda =0$$

But is it true that nilpotent matrix always has some eigenvalue? And why is it necessary that the characteristic polynomial of a nilpotent matrix is in $x^n$ form?

From what I have learned the characteristic polynomial contains all and only eigenvalues of $A$, but how can I be sure that it has any eigenvalue at all?

Answer 1

There's a very elementary proof that nilpotent matrices have both eigenvalues and eigenvectors.

Suppose that $A$ is $n \times n$ and nilpotent, and $k \le n$ is the least positive integer such that $A^k = 0$. If $k = 1$, then $A$ is the zero matrix; all non-zero vectors are eigenvectors with eigenvalue $0$, as $Av = 0v$ for $v \neq 0$.

Otherwise, $2 \le k \le n$. Then $A^{k-1} \neq 0$, so the nullspace of $A^{k-1}$ is not the entirety of $\Bbb{R}^n$. Pick some vector $v \in \Bbb{R}^n \setminus \operatorname{null} A^{k-1}$. Let $w = A^{k-1}v$, which must be non-zero.

Then $Aw = AA^{k-1}v = A^kv = 0 = 0w$. This makes $w$ an eigenvector, corresponding to eigenvalue $0$.

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More generally (and less elementary), every complex matrix has a complex (i.e. in $\Bbb{C}$, but not necessarily not in $\Bbb{R}$) eigenvalue, and non-trivial complex eigenspace to go with it. So, in general, you may take it as gospel that an eigenvalue/eigenvector pair exists. As you show, there can only be one such eigenvalue in the case of a nilpotent matrix.

Answer 2

Here's a completely elementary proof. Let $A:V\to V$ be a linear transformation, and suppose that $A^n=0$ for some $n\in \mathbb{N}$. By the rank-nullity theorem, either $A$ has a non-zero kernel or $A$ is a bijection. If $A$ is a bijection then all powers of $A$ are bijections. But $A^n=0$, a contradiction. Thus $\ker A\neq 0$. Clearly the $0$-eigenspace (i.e., the collection of all eigenvectors with eigenvalue $0$) is the kernel. And we are done.