A symmetric positive definite, $\Vert x \Vert = \sqrt{x^tAx}$ norm? Equivalent to euclidean norm?

Question

Let $A \in \mathbb{R}^{n,n}$ be a symmetric positive definite matrix. We define $\Vert x \Vert_A = \sqrt{x^tAx}$. I want to prove that is a norm. Moreover I want to prove that it is equivalent to the euclidean norm.

I know that $\Vert x \Vert_A$ is always positive and it is 0 if and only only when x = 0, so the first property is satisfied.

Moreover, it is also immediate that $\Vert \alpha x \Vert_A $ = $\vert \alpha \vert \Vert x \Vert_A$.

However, I am struggling to prove that $$ \Vert v_1 + v_2 \Vert \leq \Vert v_1 \Vert + \Vert v_2 \Vert$$

I think I should like the proof of Schwarz inequality, but I do not see how to do that here.

Supposing it is a norm, how do I prove the equivalence with the Euclidean metric? I know that I have to find two positive numbers m,M such that: $$ m\Vert x \Vert \leq \Vert x \Vert_A \leq M\Vert x \Vert $$

Any tips?

Answer 1

We want to prove that

$$\|v_1+v_2 \|_A \le \|v_1\|_A+\|v_2\|_A$$

$$(v_1+v_2)^TA(v_1+v_2) \le v_1^TAv_1+v_2^TAv_2+2\sqrt{(v_1^TAv_1)(v_2^TAv_2)}$$ which is equivalent to $$(v_1^TAv_2 )^2\le (v_1^TAv_1)(v_2^TAv_2)$$

Let $A = LL^T$ be the cholesky factoriazation, the statement become

$$((L^Tv_1)^T(L^Tv_2))^2 \le ((L^Tv_1)^T(L^Tv_1))((L^Tv_2)^T(L^Tv_2))$$

Let $u=L^Tv_1$ and $v=L^Tv_2$ and use Cauchy-Schwarz should help.

As for equivalence property, for $x$ non-zero,we have $$\lambda_{\min} \le \frac{x^TAx}{x^Tx}\le \lambda_{\max}$$

Answer 2

Since $A$ is real and symmetric, there is an orthogonal $U$ such that $U^T A U = \Lambda = \operatorname{diag}(\lambda_1,..., \lambda_n)$.

Since $A >0$ then $\lambda_n > 0$.

Let $\sqrt{\Lambda} = \operatorname{diag}(\sqrt{\lambda_1,}..., \sqrt{\lambda_n)}$, then we can write $\|x\|_A = \| \sqrt{\Lambda} U^Tx \|_2$ and since $ \sqrt{\Lambda} U^T$ is invertible, all the properties follow from those of the Euclidean norm.

Note that $\lambda_n \|x\|_2^2 \le \langle x, Ax \rangle \le \lambda_1 \|x\|_2^2$ (assuming the eigenvalues are ordered), so we see that $\sqrt{\lambda_n} \|x\|_2 \le \|x\|_A \le \sqrt{\lambda_1} \|x\|_2$

Answer 3

Another way of showing $||.||_{A}$ is a norm is just to notice that since A is symmetric and positive definite:

Let $A = LL^{T} $ be the Cholesky decomposition $$ ||x||_{A}^{2} = x^{T}Ax = x^{T}(LL^{T})x = (L^{T}x)^{T}(L^{T}x) = ||L^{T}x||_{2}^{2}$$

So $||x||_{A} = ||L^{T}x||_{2}$ and you know $ ||.||_{2} $ to be a norm.