I have a square $n \times n$ matrix $\mathbf{X}$ and a non-square $n \times m$ matrix $\mathbf{A}$. The product $\mathbf{M} = \mathbf{A}^T \mathbf{X} \mathbf{A}$ gives a $m \times m$ matrix. If I know that $\mathbf{X}$ is positive definite (or positive-semi definite), is there any way to know $\mathbf{M}$ is positive definite (or positive-semi definite).
If it is not possible, is there any constraints we can make on $\mathbf{A}$ (or maybe $\mathbf{X}$) to have this property? Any pointers will be helpful. Thanks!
Best Answer
Since $\mathbf{X}$ is positive-definite, $y^T \mathbf{X} y > 0$ for all $y\in\mathbb{R}^n, y\neq 0$. So for every $x \in \mathbb{R}^m$, we have $$ x^T (\mathbf{A}^T \mathbf{X} \mathbf{A})x = (\mathbf{A}x)^T \mathbf{X} (\mathbf{A}x) \geq 0. $$ This means $\mathbf{M}=\mathbf{A}^T \mathbf{X} \mathbf{A}$ is positive semi-definite.
If we assume that $\mathbf{A}$ has full rank, then $\mathbf{M}$ is positive-definite. Indeed, then for every non-zero $x\in\mathbb{R}^m$, $\mathbf{A}x$ is also non-zero and thus $x^T (\mathbf{A}^T \mathbf{X} \mathbf{A})x > 0$.