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Let $A,B$ be some real matrices, each $n\times n$. Given that $$rank(A) + rank(B) \le n,$$ show that there exists a real $n \times n$ matrix $C$, with $rank(C) = n$, such that $ACB = 0$.

I cannot figure out how to show that such a matrix exists. So far I have tried solving the problem using Forbenius inequality but I cannot prove that such a matrix exists. How I can approach such a problem?

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What can you say about the rank of $B$ relative to the nullity of $A$?

Note that if $U$ and $W$ are any two subspaces of the same dimension, there's a full-rank map on $n$-space that sends $U$ to $W$. Proof: Pick a basis $u_1, \ldots, u_k$ of $U$; extend it by $ a_1, \ldots, a_{n+k}$ to a basis of $n$-space. Do the same for $W$. Now send $u_i$ to $w_i$ and $a_i$ to $b_i$ for each $i$ that makes sense, and you've defined the isomorphism.

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  • $\begingroup$ I can say that rank(B) <= dim(N(A)). I cannot figure out how to use the second hint. $\endgroup$ – SebiSebi Feb 22 '16 at 19:19
  • $\begingroup$ @SebiSebi: If $ACB = 0$, what should happen to $\operatorname{Im}(B)$? $\endgroup$ – user251257 Feb 22 '16 at 19:25
  • $\begingroup$ @user251257 $Im(B) \subseteq N(AC) $. $\endgroup$ – SebiSebi Feb 22 '16 at 19:30
  • $\begingroup$ @SebiSebi: Use the hint in the answer to achieve this... $\endgroup$ – user251257 Feb 22 '16 at 19:31
  • $\begingroup$ ..and $N(AC) = N(A)$, because $C$ is an isomorphism. So you need for $C$ to transform the subspace $Im(B)$ into $N(A)$. $\endgroup$ – John Hughes Feb 22 '16 at 19:32

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