Show that if we have a matrix $A$, which can be written as $A=DD^T$ for some matrix $D$, then...

Question

we have $\bar{x}^TA\bar{x}\geq 0$ for any column vector $\bar{x}$.

I do not really know where to start, but I do know that I can somehow show the scalar $\bar{x}^TA\bar{x}$ is equal to $||D^T\bar{x}||^2$ and that would be enough. Unfortunately, I cannot see how.

Answer 1

Let's say $x \in \mathbb{R}^n$. Since $A = D D^T$, it's easy to see that $A$ must be square matrix so $A \in \mathbb{R}^{n \times n}$. Let $D \in \mathbb{R}^{n \times m}$.

\begin{align} x^T A x &= x^T D D^T x\\ &\stackrel{(1)}{=} (D^T x)^T (D^T x)\\ &\stackrel{(2)}{=} ||D^T x||^2\\ &\geq 0 \end{align}

Equality $(1)$ follows by the standard result $(AB)^T = B^T A^T$ and $(B^T)^T = B$ for any matrices $A$ and $B$.

Equality $(2)$ follows because $D^T x \in \mathbb{R}^m$, i.e. $D^T x$ is a column vector. For any column vector $v$, we have $||v||^2 = v^T v$

Answer 2

\begin{align*} ||D^T\bar{x}||^2 &=\langle D^T\bar{x} , D^T\bar{x}\rangle\\ &= (D^T\bar{x})^TD^T\bar{x}\\ &=\bar{x}^TDD^T\bar{x}\\ &=\bar{x}^TA\bar{x} \end{align*}