Hermitian matrix $k'{th}$ power eigenvectors/values.

Question

This is an undergraduate, coursework exercise:

Suppose that B is an $n\times n$ Hermitian matrix. Suppose $B^{k}X = 0$, for a $k \in N, k\geq 1$. Prove that $BX = 0$. Thank you in advance, for your help!

Answer 1

It suffices to show that

Lemma: For any $v$, $B^2v = 0 \implies Bv=0$.

With this, if we put $v=B^{k-2}X$, $B^kX=0 \implies B^{k-1}X=0$, and we can repeat this until we get to $B^2X$, at which point one final application with $v=X$ gives us $BX=0$.

To prove the lemma, if $B^2v=0$, we have by the Hermitianness (is that the word?) of $B$ that $$ 0 = \langle v , B^2 v \rangle = \langle Bv,Bv \rangle, $$ and the positive-definiteness of the inner product implies that this holds if and only if $Bv=0$. $\square$