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Let $A$ be a $n \times k$ matrix, $A\neq 0$. We know that $(A^T A)$ is positive semi-definite as:

$v^T (A^T A) v = (Av)^T (Av) = ||Av||^2 \geq 0 $

and furthermore, positive definite if $Av \neq 0 \; \forall v \neq 0$ .

Due to $A$ not necessarily being a square matrix the usual arguments of determinants and inverse are out the window, but we can still say that if $Av = 0$ for some $v = (v_1, \cdots, v_k) \neq (0,\cdots,0)$, we must have

$v_1 \begin{pmatrix} A_{11}\\\vdots\\A_{n1} \end{pmatrix} + \cdots + v_k \begin{pmatrix} A_{1k}\\\vdots\\A_{nk} \end{pmatrix} = 0 $

ie. the columns of $A$ are linearly dependent.

So if the columns of $A$ are linearly independent, ie. $A$ has full (column?) rank, this can never happen. Is this a necessary and sufficient condition for $A^T A$ to be positive definite?

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  • $\begingroup$ The indices in your column equation are off. $\endgroup$
    – balddraz
    Nov 15, 2020 at 7:13
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    $\begingroup$ what if it is the case that $A=0$ $\endgroup$
    – user844292
    Nov 15, 2020 at 7:32

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Let $A$ be an $m\times n$ matrix, then the symmetric matrix $A^{T}A$ is automatically positive semi-definite $$ x^{T}A^{T}Ax=(Ax)^{T}Ax=\|Ax\|^{2}\geq 0 $$ Now take note of the following three cases :

$1)$ If $m=n$, then $A$ is a square matrix and $A^{T}A$ is for sure a square matrix and thus $A^{T}A$ is postive-definite if and only if $A$ is of full rank.

$2)$ If $m< n$ then $A^{T}A$ is not-full rank in-fact the rank is at most $m$. Therefore, it is only positive semi-definite.

$3)$ If $m>n$ and this is the case that you should be careful with because the rank of $A$ is at most $n$ and thus it may be positive definite or positive semi-definite.

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