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If $A$ is an $n\times m$ matrix, is the formula $im(A)=im(AA^T)$ necessarily true? Explain.

I believe this to be true when $n=m$ but am unable to prove if it's true where $n\neq m$. I also don't understand how to prove either way with a formula.

any help would be appreciated.

Solumilkyu
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Lanous
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  • http://math.stackexchange.com/questions/517886/column-space-of-aa-is-equal-to-column-of-a – thanasissdr Apr 27 '16 at 06:19
  • Marked as duplicate because $A$ and $A^T$ have the same rank (so it makes no difference to say the rank of a product of two transposed matrices equals the rank of the first or of the second), and the inclusion $\def\im{\operatorname{im}}\im(AA^T)\subseteq\im(A)$ is immediate, so equality of their ranks suffices. – Marc van Leeuwen Apr 27 '16 at 14:47
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    Also note the result is not true in the stated generality: it is true for real matrices (or matrices over an ordered field), but not for complex matrices (take $A=(1~~\mathbf i)$). – Marc van Leeuwen Apr 27 '16 at 14:50
  • @MarcvanLeeuwen It may be quite apparent to you that this is a duplicate but to me and most of my classmates it does not make sense. We don't understand your answer. What if the rank is not equal? The assumption is that the rank may be unequal. – Lanous Apr 27 '16 at 18:52
  • If the ranks are not equal, then neither the images, which you want to prove would be failed. So the fact that ranks are equal is an additional property of your original problem. – Solumilkyu Apr 27 '16 at 21:01
  • OK, so here a bit more detail. First, by definition of rank (dimension of the image), having $\im(A)=\im(AA^T)$ will imply that $A$ and $AA^T$ have the same rank, but the latter also implies $\im(A)=\im(AA^T)$, since $\im(AA^T)$ is then subspace of $\im(A)$ of the same dimension, which forces it to be all of $\im(A)$. So proving $\im(A)=\im(AA^T)$ is equivalent to proving $\def\rk{\operatorname{rank}}\rk(AA^T)=\rk(A)$. Second it is well known that $\rk(A)=\rk(A^T)$, so it is also equivalent to proving $\rk(AA^T)=\rk(A^T)$. But that is done in the linked question (swapping names of $A,A^T$). – Marc van Leeuwen Apr 28 '16 at 05:25
  • Concerning $\rk(A)=\rk(A^T)$, see this question and also this question asking for an intuitive explanation. – Marc van Leeuwen Apr 28 '16 at 07:45
  • Thank you for elaborating, the new information makes it very clear. – Lanous May 01 '16 at 02:47

1 Answers1

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The relation ${\rm im}{(AA^T)}\subseteq{\rm im}{(A)}$ is trivial, so we only show that ${\rm im}{(A)}\subseteq{\rm im}{(AA^T)}$. Before starting, we claim that $\ker(A)^\perp\subseteq {\rm im}(A^T).$ If otherwise $x\notin{\rm im}(A^T)$, then there exists $x'\in{\rm im}(A^T)^\perp$ such that $ x\cdot x'\ne0$. In fact, we have $x'\in\ker (A)$ because $A^TAx'\in{\rm im}(A^T)$, which implies $$Ax'\cdot Ax'=(x')^TA^TAx'=x'\cdot A^TAx'=0\quad\Longrightarrow\quad Ax'={\it 0}.$$ Thus $x\notin \ker (A)^\perp$, and the claim is proved. Now, given $y\in{\rm im}{(A)}$, there exists $x\in\mathbb{R}^m$ such that $y=Ax$. Also, we may write $x=x_1+x_2$, where $x_1\in\ker{(A)}$ and $x_2\in\ker{(A)}^\perp$. By the claim, $x_2=A^Tx_2'$ for some $x_2'\in\mathbb{R}^n$. Therefore \begin{align} y =A(x_1+x_2) =A(x_1+A^Tx_2') =Ax_1+AA^Tx_2' =AA^Tx_2', \end{align} and hence $y\in{\rm im}(AA^T)$.

Solumilkyu
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