Assume we have $A \in \mathbb{R}^{m \times n}$, $rg(A) = m$. Will $\det(AA^T)$ be non-zero? Same question is for $\det(\Phi^T\Phi)$, where $\Phi \in \mathbb{R}^{n \times m}, rg(\Phi) = m$.
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What exactly does "rg" stand for? – Ben Grossmann Mar 12 '21 at 18:46
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rank of matrix A – Blinkop Mar 12 '21 at 18:47
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Presumably $m \le n$? – Theo Bendit Mar 12 '21 at 18:49
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I think $rank(A) = m$ stands for $m <= n$ – Blinkop Mar 12 '21 at 18:52
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Yes, it is true: if $A$ is $m \times n$ with rank $m$, then $\det(AA^T)$ will be non-zero. Similarly, if $A$ is $m \times n$ with rank $n$, then $\det(A^TA)$ will be non-zero.
There are a few ways you could go about proving this. One is to use the fact that $$ \operatorname{rank}(A) = \operatorname{rank}(A^TA) = \operatorname{rank}(AA^T), $$ and a square matrix will have non-zero determinant if and only if its rank is equal to its size. Alternatively, you could use the Cauchy Binet formula to express either determinant as a sum of squares.
Ben Grossmann
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I am sorry, but I cannot understand the $rank(A^TA) = rank(AA^T)$ transition. Could you please explain, because I have a feeling that I miss something fundamental about this thing? – Blinkop Mar 12 '21 at 19:00
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@Blinkop It is simply a fact that $\operatorname{rank}(A) = \operatorname{rank}(A^TA)$ and $\operatorname{rank}(A) = \operatorname{rank}(AA^T)$. The fact that this holds in the case that $A$ has full rank is proved here, but if you're interested I could either write or point you to a more general proof – Ben Grossmann Mar 12 '21 at 19:03
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