Triangle Inequality of the "A"-norm

Question

Suppose $A$ is an $n\times n$, symmetric, positive definite matrix, i.e. $(Ax,x) > 0$ for all $0 \neq x \in \mathbb{R}^{n}$ and $A^{T} = A$. Propose the following norm, called the "A"-norm, by $$ \|x||_{A} := (Ax,x)^{1/2}.$$ Trivially, this satisfies nonnegativity of the norm as well as scalar multiplication. How may it be shown that the triangle inequality, i.e. $\|x+y\|_{A} \leqslant \|x\|_{A} + \|y\|_{A}$, is also satisfied?

Answer 1

Thanks to the hypothesis on $A$, there exists $B$ symmetric and positive-definite such that $A=B^2=B^TB$. Therefore $$ \|x\|_A^2=(Ax,x)=(B^T Bx,x)=(Bx,Bx)=\|Bx\|^2, $$ where $\|\cdot\|$ is the usual euclidean norm. Since the euclidean norm satisfies the triangle inequality, so doeas the $A$-norm $\|\cdot\|_A$.

Addendum. The first statement can be quite easily proved by using the diagonalization theorem: any symmetric matrix $S$ can be diagonalized by means of an ortogonal matrix $R$: $$ \exists R:\quad R^T=R^{-1},\quad R^TSR=D \quad (\textrm{or } S=R D R^T) $$ with $D$ diagonal: $D=\textrm{diag}(\lambda_1,\ldots,\lambda_n)$.

If $S$ is non-negative definite, than $\lambda_i\geq 0$. Let $D'=\textrm{diag}(\lambda_1^{1/2},\ldots,\lambda_n^{1/2})$ and $S':=R D' R^T $. Then $S'$ is symmetric (trivial) and $S=S'^2 $; in fact, $$ S'S'=(R D' R^T)(R D' R^T)=R D' (R^T R ) D' R^T=R D' D' R^T=R DR^T=S $$