What could be the rank of a matrix multiplied by its transpose ?
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
Updated on June 11, 2020Comments
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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!
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fgp over 9 yearsIt certainly isn't $n$, at most it can be $m$.
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