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Let $A$ a $n\times n$ matrix such that $\operatorname{rank}(A)=m$, can we say something about $\operatorname{rank}(AA^t)$?

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Please translate it to English: se pode afirmar algo sobre rank(AA^t)? –  hjpotter92 Mar 11 '13 at 18:24
    
@BackinaFlash: Can something be said about. –  Brian M. Scott Mar 11 '13 at 18:24
    
@BrianM.Scott thank's –  Henfe Mar 11 '13 at 18:25
    
You’re welcome. Did you mean $\operatorname{rank}(A)=n$ (instead of $m$) in the body of the question? –  Brian M. Scott Mar 11 '13 at 18:27
    
@BrianM.Scott Exactly, rank(A)=m –  Henfe Mar 11 '13 at 18:29

2 Answers 2

up vote 6 down vote accepted

It's true in general that $rank(A^tA)=rank(A)$. To prove that, it clearly suffices to show that $kerA^tA=ker A$, since $rank(T)=n-dim(ker T)$ for every $n\times n$ matrix $T$ (where we denote $ker T=\left\{x\in \mathbb{R}^n: Tx=0\right\}$).

Obviously, $ker A\subseteq ker A^tA$. Conversely, if $x\in ker A^tA$, i.e., $A^tAx=0$, then we can multiply that equation by $x^t$ to the left and obtain $$<Ax,Ax>=(Ax)^tAx=x^tA^tAx=x^t0=0$$ Since the inner product is positive-definite, we have $Ax=0$ if $A^tAx=0$.

Therefore, $rank A^tA=rank A(=n-dim(ker A))$. If we apply that result to $A^t$ in place of $A$, we obtain $$rank(AA^t)=rank((A^t)^tA^t)=rank(A^t)=rank(A)=m$$

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For any two square matrices $A$ and $B$, we have $\operatorname{rank}(AB)\ge\operatorname{rank}(A)+\operatorname{rank}(B)-n$. Hence in your case, we have, as pointed out by Maisam Hedyelloo in his (now deleted) answer, \begin{equation} \operatorname{rank}(AA^T)\ge2\operatorname{rank}(A)-n=2m-n.\tag{1} \end{equation} Also, in general we have $\operatorname{rank}(AB)\le \min(\operatorname{rank}(A),\operatorname{rank}(B))$. Hence in your case, \begin{equation} \operatorname{rank}(AA^T)\le\operatorname{rank}(A)=m.\tag{2} \end{equation} Over $\mathbb{R}$, equality in $(2)$ always holds. In fact, if $v\in\ker(AA^T)$, then $\|A^Tv\|^2=v^TAA^Tv=0$ and in turn $A^Tv=0$. Hence $\ker(AA^T)\subseteq\ker(A^T)$. Since inclusion in the reverse direction is always true, we conclude that $\ker(AA^T)=\ker(A)$. By the rank-nullity theorem, we get $\operatorname{rank}(AA^T)=\operatorname{rank}(A)$.

For other fields, strict inequality in $(2)$ may occur. For example, consider $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ over $\mathbb{F}_2$. Then $AA^T=0$ (because $1+1=0$) and hence $\operatorname{rank}(AA^T)=0<1=\operatorname{rank}(A)$.

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