Show $\| X \| = \sqrt{X^* A X}$ is a norm

Question

I'm trying to show that, given a positive definite matrix $A\in \mathcal{M_n}(\mathbb{C})$, the function

$$\begin{array}{cccl} \lVert \cdot \rVert: &\mathbb{C}^n &\longrightarrow &\mathbb{R} \\\ &X &\longmapsto & \lVert X \rVert = \sqrt{X^* A X} \end{array}$$

is a norm.

I've proved all properties except $\| X+Y \| \leq \| X \|+\| Y \| \ \forall \ X,Y \in \mathbb{C}^n$.

$$\| X+Y \|= \sqrt{(X+Y)^* A (X+Y)}=\sqrt{(X^*+Y^*) A (X+Y)}=\sqrt{X^*AX+X^*AY+Y^*AX+Y^*AY}$$

$$\| X\| +\|Y \|= \sqrt{X^* A X} + \sqrt{Y^* A Y}$$

I don't know how to arrive to the inequality from here.

Answer 1

You can instead show that $$ \langle x,y\rangle_A:=x^*Ay,\quad x,y\in{\bf C}^n, $$ is an inner product, which implies that $x\mapsto\sqrt{\langle x,x\rangle_A}$ defines a norm.

Essentially, the triangle inequality is a consequence of the Cauchy-Schwarz inequality. Note that in your calculation, taking the square makes things easier: $$ \|X+Y\|^2= (X+Y)^* A (X+Y) =(X^*+Y^*) A (X+Y) =X^*AX+X^*AY+Y^*AX+Y^*AY, $$ $$ (\| X\| +\|Y \|)^2 = X^* A X + Y^* A Y +2\|X\|\cdot\|Y\|. $$

Now to show the triangle inequality, it suffices to show that $$ 2\|X\|\cdot\|Y\|\geqslant X^*AY+Y^*AX. $$ But since $A$ is Hermitian, $$ X^*AY+Y^*AX=X^*AY+X^*AY=2X^*AY. $$ So you want to show $$ \|X\|\cdot\|Y\|\geqslant X^*AY, $$ which is a consequence of the Cauchy-Schwarz inequality.

Answer 2

You can get an eigenbasis diagonalization of $A$ as $A = Q \lambda Q^*$

Do the change of variables $\hat X = \lambda^{-1/2} Q^*X$.

$$||X + Y|| = ||\hat X + \hat Y||_2 \le ||\hat X||_2 + ||\hat Y||_2 = ||X|| + ||Y||$$