What could be the rank of a matrix multiplied by its transpose ?

Question

Let $A$ be a full rank $m×n$ matrix $(m<n)$, i.e. $\operatorname{rank}(A)=m$.

Can the rank of $A'A$ be $n$? Under what condition would this hold?

Thanks!

Answer 1

An $m \times n$ matrix can be identified with a linear mapping from $\mathbb{K}^n$ to $\mathbb{K}^m$. So what you do is to map from $\mathbb{K}^n$ to $\mathbb{K}^m$ and then back from $\mathbb{K}^m$ to $\mathbb{K}^n$. But $m<n$, so the second mapping cannot be injective, that is, its kernel includes more then $\vec{0}$. From this it follows that also if you combine the two mappings, some elements of $\mathbb{K}^n$ that are not $\vec{0}$ will map into the kernel of the second mapping and from there to $\vec{0}$, so all in all it cannot be injective (nor bijective). Therefore, also the product matrix cannot have full rank.

Answer 2

According to http://en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties for A with real entries: $$\operatorname{rank}(A^T A) = \operatorname{rank}(A A^T) = \operatorname{rank}(A) = \operatorname{rank}(A^T)$$

Answer 3

No. Observe that for any matrix A of size $m \times n$, $\mbox{Rank}({\bf{A}}) \leq \min(m,n)$.

So, assuming that $m < n$ and A is full rank, then $\min(n,m)=m$, and $\mbox{Rank}({\bf{A}}) = m \neq n$.

And since $\mbox{Rank}({\bf{A}})=\mbox{Rank}({\bf{A}}^\top {\bf{A}})$ for any matrix A, then $\mbox{Rank}({\bf{A}}) = \mbox{Rank}({\bf{A}}^\top {\bf{A}}) = m \neq n$

So, $\mbox{Rank}({\bf{A}}^\top {\bf{A}}) = n$ would hold only if $n \leq m$.