TITLE: Show that if $B^2 x = 0_n$ for some vector $x \neq 0_n$, then $B$ is not invertible
QUESTION [7 upvotes]: If $B \in M_{n \times n}(\mathbb{R})$ and $B^2 x = 0_n$ for some vector $x \neq 0_n$, then $B$ is not invertible.

I get that $$\mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.

REPLY [9 votes]: Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.