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Let $A$ be $m\times n$ matrix with rank $r=\min(m,n)$. How do we show that rank$(A^T A)$ is $r$.

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  • $\begingroup$ do you mind to say what $A'$ is... $\endgroup$ – user87543 Jul 2 '14 at 13:13
  • $\begingroup$ Matrix transpose, en.wikipedia.org/wiki/Transpose $\endgroup$ – Lionville Jul 2 '14 at 13:16
  • $\begingroup$ why do you make $A^\top A$ when it is usually denoted by $A^TA$ $\endgroup$ – user87543 Jul 2 '14 at 13:21
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    $\begingroup$ math.stackexchange.com/questions/48989/… should be of some help to you.. $\endgroup$ – user87543 Jul 2 '14 at 13:22
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Hint: show that $A$ and $A^TA$ have the same nullspace. That is, $$ Ax = 0 \iff A^TAx = 0 $$ It is helpful to note that $x^T(A^TAx) = \|Ax\|^2$

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What you need is to use the Singular Value Decomposition. Then, just have a look what happens with singular values. Here is a link: http://en.wikipedia.org/wiki/Singular_value_decomposition

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    $\begingroup$ Don't you think that the SVD a bit of an overkill here? $\endgroup$ – Vedran Šego Jul 2 '14 at 13:38
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    $\begingroup$ It is, but it was the first thing that came to my mind because I used it on a different problem few hours ago. $\endgroup$ – Jan Jul 2 '14 at 13:46
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Firstly, we proof than $\mathsf{Null} A=\mathsf{Null} (A^TA)$ ($A$ is an $m\times n$ matrix):

a) $\mathsf{Null} A\subset \mathsf{Null} (A^TA)$, because $\mathsf{Null} A$ is always a subset of $\mathsf{Null}(BA)$.

b) If $x\in\mathsf{Null} (A^TA)$, then $A^T(Ax)=0$ and, so, $Ax\in\mathsf{Null}(A^T)$. In the other hand, $Ax$ is in the column space of $A$, $\mathsf{Col}A$. Then, $$ Ax\in\mathsf{Null}(A^T)\cap \mathsf{Col}A $$ But we know $\mathsf{nul}(A^T)=(\mathsf{Col}A)^\perp$, and, so, $\mathsf{Null}(A^T)\cap \mathsf{Col}A=\emptyset$. Consequently, $Ax=0$, and $x\in\mathsf{Null}A$.

Now, $$ \mathsf{rank} (A^TA)=n-\mathsf{dim} \mathsf{Null} (A^TA)= n-\mathsf{dim} \mathsf{Null} A=\mathsf{rank} A $$

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