# Show that if $B^2 x = 0_n$ for some vector $x \neq 0_n$, then $B$ is not invertible

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.

A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.

If $$B^2 x = 0$$, then $$B(Bx) = 0$$, so $$Bx \in \ker B$$.

Suppose $$Bx = 0$$, then this shows that $$B$$ has a non-trivial kernel, hence is not injective.

Suppose $$Bx \neq 0$$ above, then $$Bx$$ is a non-trivial element of $$\ker B$$, so it is not injective for $$n \geq 3$$.

Hence, either way it follows that $$B$$ is not injective.

VIA CONTRAPOSITIVE : If $$B$$ is injective, then so is $$B^2$$, but then $$B^20 = 0$$, so $$B^2x$$ cannot be zero for any other $$x$$ by injectivity. Note that we can extend this to $$B^nx = 0$$ implies $$B$$ is not injective.

If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $\det(B^2)=0$.

But $\det(B^2)=\det(B)^2$, so $\det(B)=0$ and $B$ is not invertible.

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 $x=0$.