# Matrix rank inequality

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?

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.

• I can say that rank(B) <= dim(N(A)). I cannot figure out how to use the second hint. Feb 22, 2016 at 19:19
• @SebiSebi: If $ACB = 0$, what should happen to $\operatorname{Im}(B)$? Feb 22, 2016 at 19:25
• @user251257 $Im(B) \subseteq N(AC)$. Feb 22, 2016 at 19:30
• @SebiSebi: Use the hint in the answer to achieve this... Feb 22, 2016 at 19:31
• ..and $N(AC) = N(A)$, because $C$ is an isomorphism. So you need for $C$ to transform the subspace $Im(B)$ into $N(A)$. Feb 22, 2016 at 19:32

I find it always a tiny bit easier to reason with concrete numbers. Also keeping in mind the 4 fundamental subspaces picture helps. So suppose $$n=10$$.

1) the last operation is A. Let's suppose $$rank(A) = 8$$ so that the nullspace of A has dimension $$dim(N(A)) = 2$$. Because we want the result of ABC to be 0 we need that BC ends in N(A). So this implies that the column space of BC is included in N(A). Let $$a_1,a_2$$ be a basis of $$N(A)$$ and extend it with a basis of the rowspace of A $$an_3,an_4,...,an_{10}$$ so that $$a_1,a_2,an_3,an_4,...,an_{10}$$ is a basis of $$R^{10}$$

2) the first operation is B. Let'suppose that $$rank(B) = 2$$ so that applying B makes you end in a 2-D columnspace. Let $$b_1, b_2$$ be a basis of the column space of B, and extend it to a basis of $$R^{10}$$ with $$bn_3,bn_4,...,bn_{10}$$ (a basis of the left-nullspace of B) so that $$b_1, b_2, bn_3,bn_4,...,bn_{10}$$ is a basis of $$R^{10}$$

3)So now we need to find a full-rank C that connects two 2-D subspaces of $$R^{10}$$: the column space of B and the nullspace of A. For this it is enough to have a C that sends $$b_1$$ on $$a_1$$, $$b_2$$ on $$a_2$$,$$bn_3$$ on $$an_3$$...,$$bn_{10}$$ on $$an_{10}$$. Let's stack side by side the b's in a n x n matrix $$U = \begin{pmatrix}|&|&|&|&|\\b_1&b_2&bn_3&\dots&bn_{10}\\|&|&|&|&|\end{pmatrix}$$ and the a's in a matrix $$V = \begin{pmatrix}|&|&|&|&|\\a_1&a_2&an_3&\dots&an_{10}\\|&|&|&|&|\end{pmatrix}$$. We are shooting for a matrix $$C$$ such that $$CU=V$$ . U is made of 10 independent columns so it is invertible so $$C = CUU^{-1}=VU^{-1}$$. Because $$V$$ and $$U^{-1}$$ have zero nullspace so has C (*) and hence rank(C) = 10.

Hopefully you can easily generalize.

(*) If A and B have zero nullspace, then $$ABx = 0$$ implies $$Bx=0$$ because nullspace of A is 0; implies $$x = 0$$ because nullspace of B is zero: So nullspace of AB is zero.