Prove that if $A$ and $B$ are $n\times n$ matrices, $A$ is invertible and $BAB=0$ then ${\rm rank}(B) \leq n/2$

Question

Let $A, B$ are $n\times n$ matrices and $A$ is invertible. I need to show that if $BAB=0$ then ${\rm rank}(B) \leq \frac{n}{2}$.

How can I do it?

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I can prove that

$$0 = {\rm rank}(BAB) \leq {\rm rank}(BA) = {\rm rank}(B) \leq {\rm rank}(A) = n$$

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Edit: this problem assuming no knowledge about kernel

Answer 1

What is the dimension of the kernel of B?

$BAB$ maps every vector in an $n$ dimensional space to $\mathbb 0.$ $A$ is full rank, so all of this compression is happening from the $B$ matrices. Each $B$ matrix must then be mapping at least half the vectors in the original space to $\mathbb 0$?

$\text {Nullity} (BAB) \le \text {Nullity} (B) + \text {Nullity} (A) + \text {Nullity} (B)$