# Rank of the product of two full rank matrices

I have searched for the above topic and found some results, but the answer I am looking for is not found anywhere. Here is my question:

Given $A_{m \times n}$ matrix with rank $m$, and $B_{n \times p}$ matrix with rank $p$, where $n > p \geq m$. I know that $$\operatorname{rank}(AB) \leq \min\left(\operatorname{rank}(A),\operatorname{rank}(B)\right)$$ What I want to know is if this expression holds for equality. I.e is this expression $$\operatorname{rank}(AB) = \min\left(\operatorname{rank}(A),\operatorname{rank}(B)\right) = m$$ correct? If yes, how can it be proved?

• What have you tried yourself to prove it? Have you tried to find matrices $A$ and $B$ that serve as a counterexample? – Mike Pierce Aug 28 '15 at 14:23
• I did create Matrix A and B and find the rank of the Product, it gives me the smaller of rank(A) and rank(B). which means rank(AB) = min(rank(A),rank(B)). Now what i don't know is weather this is true in general, as what i saw in general is the inequality, but don't know if the equality hold to. – user5275281 Aug 28 '15 at 15:59

if we take A and B be 2 non-zero matrices s.t AB= zero matrix then rank A,rank B >0 but rank of AB = 0 so rank AB is not equal to min{rank A, rank B}

• It seems to me that in the question, $A$ should have rank $n$ and $B$ rank $p$... – Peter Franek Aug 28 '15 at 14:31
• Hi Peter Franek, i am sorry for the confusion in my initial post post. The rank of A should be m, not n and the rank of B is p. Also n > p >= m. – user5275281 Aug 28 '15 at 16:15

(This is a reply to an earlier version of the question with $rank(A)=n$, $rank(B)=p$).

Yes, it holds. The assuptions imply $m\geq n\geq p$ and so you can find a submatrix $A'$ of $A$ that contains $n$ rows of $A$ and hence has size $n\times n$ and is regular. The columns of $B$ are independent and so are the columns of $A'B$. So indeed, $$\mathrm{rank} AB=\mathrm{rank} A'B=p=\min\{n,p\}.$$

But as Rupsa pointed out, if the matrices $A,B$ have smaller then the full column rank, then equality doesn't hold in general, such as in $(1,0){0\choose 1}=(0)$.

• Thank Peter, look great, but my assumption is $n > p \geq m$, not as you mentioned above. – user5275281 Aug 28 '15 at 16:24
• Ah, I see. But then all is clear, as the counter-example above applies to your situation ($2>1\geq 1$). – Peter Franek Aug 28 '15 at 16:29