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Let $A$ a real matrix of $n\times n$.

I want to prove that if $A$ is not invertible, then there exists $B\in M_n(\mathbb{R})\setminus\{0\}$ such that $AB=0$.

How can we find such a matrix $B$?

Thanks :)

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Here's a hint: if $A$ is not invertible, it has nontrivial nullspace, meaning that there is some nonzero vector $x$ such that $Ax = 0$. How can you relate this to $B$?

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If $A$ is not invertible, then there exists a nonzero vector $X$ such that $AX = 0$. Take $B = \pmatrix{X & 0}$, where $0$ consists of $n-1$ columns of zeros.

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