Eigenvalues of a nilpotent matrix can only be $0$

Question

Prove that the eigenvalues for a square Nilpotent matrix A can only be $0$.

Definition of nilpotent

A $^n$=$0$

n is a positive whole integer

Answer 1

One can also work directly, without the apparatus of the minimal polynomial:

If $V$ is a vector space over the field $\Bbb F$, then by definition an eigenvalue satisfies

$Ax = \lambda x, \; \;, 0 \ne x \in V; \lambda \in \Bbb F, \tag{1}$

then

$A^2 x = A(Ax) = A(\lambda x) = \lambda (Ax) = \lambda^2 x, \tag{2}$

and so if we assume

$A^k x = \lambda^k x \tag{3}$

then we have

$A^{k + 1} x = A(A^k x) = A(\lambda^k x) = \lambda^k Ax = \lambda^{k + 1} x; \tag{4}$

taking (1)-(2) as base cases, we have thus completed a simple inductive proof that

$A^n x = \lambda^n x \tag{5}$

for all integers $n \ge 1$. Now if

$A^n = 0, \tag{6}$

then an eigenvalue $\lambda$, satisftying (1), also satisfies

$\lambda^n x = A^n x = 0, \tag{7}$

and since $x \ne 0$ we must have

$\lambda^n = 0, \tag{8}$

whence

$\lambda = 0. \tag{9}$.

QED.

In response to FemptoComm's comment to Timbuc's answer, the chances are diminished by the fact that nonzero nilpotent matrices $A$ are can't be diagonalized, and that $\det A = 0$ only allows us to conclude that some eigenvalue is zero, not all eigenvalues. I leave the proofs of these well-known facts to may readers.

Disclaimer: I read Timbuc's comment to his own answer after writing this up.

Hope this helps. Holiday Cheers,

and as ever,

Fiat Lux!!!

Answer 2

Hint:

$$A^n=0\implies A\;\;\text{is a root of}\;\;x^n\implies\;\text{its minimal polynomial divides}\;\;x^n$$