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

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Solution 1

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

Solution 2

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 $n \leq m$.

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Jingjings
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Updated on June 11, 2020

Comments

  • Jingjings
    Jingjings over 3 years

    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!

    • fgp
      fgp over 9 years
      It certainly isn't $n$, at most it can be $m$.