# When is $A^T A$ a positive definite matrix for non square $A$?

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?

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

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.