# Full-rank condition for product of two matrices

Given two matrices $A_{m\times n}$ and $B_{n\times p}$, what is the sufficient and necessary condition for $AB$ to have full rank?

I know $r(AB)=r(B)-\dim N(A) \cap R(B)$, so is it true the above iff $\dim N(A)\cap R(B)=0$? But seems incorrect if $m\lt p$. Please help.

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Related post: 93989. See Robert Israel's answer in that post. – Srivatsan Dec 26 '11 at 22:20
@Srivatsan: I don't see how the answer can be directly translated to my question, unless you show explicitly. Thanks. – littleEinstein Dec 26 '11 at 22:26
Seems to me that if $n$ is the smallest of the three numbers then full rank is impossible. – Gerry Myerson Dec 27 '11 at 0:11

## 2 Answers

Assume we want to multiply two matrices $A$ and $B$, and we want to calculate the rank. All solutions that satisfy $ABx=0$ are in the NULL space of $AB$. If $Bx=0$, then the equation is satisfied and therefore everything in the NULL space of $B$ is in NULL space of $AB$. Now $A$ comes along, if there is something in the NULL space of $A$, that is, $Ax=0$, and that combination is spanned by column space of $B$, then we have a vector $z=Bx$ such that it is in NULL space of $A$ therefore reducing the rank. Conclusion: True!

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Somewhat expanding the comments, one can say the following. The rank of the product of a $m\times n$ and a $n\times p$ matrix can never exceed $n$ (nor can it exceed $m$ or $p$, but that is obvious from the size of the product). This is because, if you want to find $r$ (for rank) vectors in the source space (of dimension $p$) whose images by the product of the matrices are linearly independent, then their images by the first matrix applied ($B$) must also be linearly independent, and this can only be the case if the dimension if the intermediate space permits it: $r\leq n$. So if $n<\min(m,p)$ then the product can never have full rank.

If $\min(m,p)\leq n\leq \max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. This condition of full rank factors is not a necessary one though: if $m\geq n>p$ then $A$ might have a nonzero kernel, as long as it forms a direct sum with the image of $B$, while if $m<n\leq p$ then the image of $B$ need not be the whole space, as long as its sum with the kernel of $A$ fills that space.

When $n>\max(m,p)$ all one can say in general, knowing just the ranks of $A$ and $B$, is the usual necessary condition either that $B$ be injective (if $m\geq p$) or that $A$ be surjective (if $m\leq p$) in order for $AB$ to be injective respectively surjective, that is, full rank. Again the precise condition for $AB$ to be injective is $\ker A\oplus\mathop{\textrm{im}}B$ (this gives "full rank" when $m\geq p$), and for $AB$ to be surjective is $\ker A+\mathop{\textrm{im}}B=\mathbf R^n$ (this gives "full rank" when $m\leq p$).

It may be interesting to note that one can define the rank of a $m\times p$ matrix $C$ as the minimal value $r$ such that there exists a decomposition $C=AB$ with $A$ of size $m\times r$ and $B$ of size $r\times p$. It does not immediately give a good method to compute the rank, but lets you understand immediately that you cannot have full rank in your question unless $n\geq\min(m,p)$.

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