# Positive Definite Matrix Norm

I am attempting to prove that a norm where for any $$(k \times 1)$$ vector x with $$(1 \times k)$$ transpose y:

$$\|x\|=\sqrt{yVx}$$

for some $$(k \times k)$$ positive definite (and symmetric) matrix $$V$$, is in fact a norm. Positive Definiteness and Homogeneity are quite trivial to prove, but I'm struggling with the triangle inequality.

Is it possible to prove the triangle inequality (easily?), or should I just prove this is an inner product instead(which I believe is a simpler task) and use that to say that it is a norm?

Thank you!

• Is it possible that you also need $V$ to be symmetric? Otherwise, you do not even have the inner product you talk about. – michalOut Nov 4 '18 at 21:48
• V must also be symmetric, however I believe that is a requirement for it to be positive definite in the first place. (Which I why I didn't mention it) – vulcan583 Nov 4 '18 at 22:10
• Did you mean that $\| x \| = \sqrt{x^\top V x}$ ? This is the usual definition of the induced norm you describe. – VHarisop Nov 4 '18 at 22:19
• If you define positive definite by property: $x^TVx > 0$ for all $x \neq 0$ (which is usual), then no. For example, consider $A = \begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}$. For any $x=[x_1, x_2]^T \neq 0$ one has $x^TAx = [x_1, x_2] [x_1 + x_2, -x_1 + x_2]^T = x_1^2 + x_2^2 >0$ and at the same time $A$ is not symmetric. – michalOut Nov 4 '18 at 22:52
• yes VHarisop, there is a squareroot, I wasnt sure it was relevant. – vulcan583 Nov 5 '18 at 0:27

Denote $$\| x \|_V = \sqrt{x^\top V x}$$ and $$\| x \|_2$$ for the usual vector norm. We can write

$$\| x + y \|^2_V = (x+ y)^\top V (x + y) = x^\top V x + y^\top V y + x^\top V y + y^\top V x \\ = \|x\|_V^2 + \|y\|_V^2 + 2 x^\top V y.$$ Now notice that since $$V$$ is symmetric positive definite, it admits a square root, i.e. $$V = V^{1/2} V^{1/2}$$. Using this and the Cauchy-Schwarz inequality one may write

$$x^\top V y = x^\top V^{1/2} V^{1/2} y = (V^{1/2} x)^\top (V^{1/2} y) \leq \| V^{1/2} x\|_2 \| V^{1/2} y \|_2$$

However, we know that $$\| z \|_2 = \sqrt{z^\top z}$$, so replacing in the above expression we get

$$2 x^\top V y \leq \sqrt{x^\top V^{1/2} V^{1/2} x} \sqrt{y^\top V^{1/2} V^{1/2} y} = \sqrt{x^\top V x} \sqrt{y^\top V y} = 2 \| x \|_V \| y \|_V$$

Finally, this gives us

$$\| x + y \|_V^2 \leq \|x \|_V^2 + \| y \|_V^2 + 2 \| x \|_V \| y \|_V = (\| x \|_V + \| y \|_V)^2$$ so taking away the square gives us $$\| x + y \|_V \leq \| x \|_V + \| y \|_V$$, as required.