Symmetric Positive Definite Matrices and Norms

Question

Suppose $A \in \mathbf{R}^{n\times n}$ is symmetric positive definite. Show that $||x||_A = \sqrt{x^TAx}$ defines a norm for the vector $x$.

Why are the first two properties of norms obvious? That is, why can:

• $||x||_A \geq 0, \forall \space x; ||x||_A = 0 \leftrightarrow x = 0$
• $||ax|| = |a|||x||, \quad \forall \space x$

be easily seen? I can prove the triangle property (property 3: not listed), but does the first property follow directly from the fact $A$ is symmetric positive definite? What about the second? Thanks!

Answer 1

1. Non-negativity:

$A\succ 0$ implies that $x^\top A x > 0$ for all $x\neq 0$, so $\lVert x\rVert_A > 0$. On the other hand $x^\top A x = 0 \iff x = 0$, so $\lVert x\rVert_A \geq 0$.

2. Triangle inequality:

Using Cauchy-Schwarz $$ \lVert x + y \rVert_A^2 = \lVert x\rVert_A^2 + 2x^\top Ay + \lVert y\rVert_A^2\leq \lVert x\rVert_A^2 + 2\lVert x\rVert_A\lVert y\rVert_A + \lVert y\rVert_A^2 = (\lVert x\rVert_A + \lVert y\rVert_A)^2, $$ which implies $\lVert x + y \rVert_A\leq \lVert x\rVert_A + \lVert y\rVert_A$.

3. Homogeneity: $$ \lVert ax \rVert_A = \sqrt{(ax)^\top A(ax)} = \sqrt{a^2x^\top Ax} = \lvert a\rvert\sqrt{x^\top Ax} = \lvert a \rvert\lVert x \rVert_A $$