I already read some several posts about this question, but they didn't resolve my problem. I want to prove that $||x||=(x^TAx)^{1/2}$ is a vector norm on $R^n$.
A.) $\Vert x \Vert \ge 0, \;\; \forall x \in V; \; \; \Vert x \Vert = 0 \Leftrightarrow x = 0$;
B.) $\Vert ax \Vert = \vert a \vert \Vert x \Vert, \;\; \forall x \in V, a \in \Bbb R$;
C.) $\Vert x + y \Vert \le \Vert x \Vert + \Vert y \Vert, \;\; \forall x, y \in V$.
Verifying (A) and (B) is rather straightforward. But to check triangle inequality, here's the steps I followed:
$||x+y||^2=(x+y)^TA(x+y) = (x+y)^TLL^T(x+y)$, where $L$ is a Cholesky factor.
Then $(x+y)^TLL^T(x+y) = (x^TL + y^TL)(L^Tx+L^Ty)=x^TLL^Tx+x^TLL^Ty+y^TLL^Tx+y^TLL^Ty = ||x||^2+2x^TLL^Ty + ||y||^2.$
I want $||x||^2+2x^TLL^Ty + ||y||^2 \le ||x||^2+2||x||||y|| + ||y||^2 = (||x||+||y||)^2$ to finish the proof (or something along this line), how can I prove $2x^TLL^Ty \le 2||x||||y|| $?
We want to show that
$$(x^TL)(L^Ty) \leq \|x\|\|y\|=(x^TLL^Tx)^\frac12(y^TLL^Ty)^\frac12$$
Squaring both sides, this is just Cauchy-Schwarz inequality.
$$((L^Tx)^T(L^Ty))^2 \le [(L^Tx)^T(L^Tx)][(L^Ty)^T(L^Ty)]$$
If you let $u=L^Tx$ and $v=L^Ty$ and $\langle u, v \rangle = u^Tv$, it is
$$\langle u, v \rangle^2 \le \langle u,u \rangle \langle v,v \rangle$$