Let A be an $n\times p$ matrix with rank $p$. I want to prove that $X = A^TA$ is a positive definite matrix.
I know that for any symmetric matrix to be positive definite, all of its eigenvalues must be strictly positive.
So I need a way to show that $X$'s eigenvalues are all positive, or perhaps any other method to proof that $X$ is positive definite.
Best Answer
As suggested above by phdmba7of12, we consider the quadratic form
$$x^T X x = x^T (A^T A)x = (Ax)^T (Ax) = (Ax, Ax) = \lVert Ax \rVert^2 \geq 0.$$
Since $A$ has rank $p$, the only solution of $Ax = 0$ is the trivial solution. Hence,
$$x^T X x = 0 \iff x = 0.$$
Hence, we have varified the definition of positive definiteness.