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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.

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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.

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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.

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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 $x=0$.

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