Let $A$ be a $5\times5$ matrix that can be written in the form $A=BC$, where $B$ is a $5 \times 4$ matrix and $C$ is a $4 \times5$ matrix. Prove that $A$ is not invertible.
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2Have you tried anything? A good start might be to see if $A$ has a non-trivial null space.... – Jan 17 '14 at 21:19
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@T.Bongers you mean that $Ax=0$ has only solution $x=0$? – user121819 Jan 17 '14 at 21:21
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No he means that there is a non zero $x$ such that $Ax=0$ – voldemort Jan 17 '14 at 21:22
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Note that the rank inequality for products give us that $rank(A) \leq rank(B)$.
Now, $B$ can have rank $4$ at most- do you see why?
So, $rank(A) < 5$ hence $A$ is not invertible
voldemort
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Note that column space of $A$ is contained in column space of $B$. So rank $A \leq rank(B)$ – voldemort Jan 17 '14 at 21:24
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This will help: http://math.stackexchange.com/questions/48989/how-to-prove-textrankab-leq-min-textranka-textrankb – voldemort Jan 17 '14 at 21:25