1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Is it possible that you also need $V$ to be symmetric? Otherwise, you do not even have the inner product you talk about. $\endgroup$ – michalOut Nov 4 '18 at 21:48
  • $\begingroup$ 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) $\endgroup$ – vulcan583 Nov 4 '18 at 22:10
  • 2
    $\begingroup$ Did you mean that $ \| x \| = \sqrt{x^\top V x} $ ? This is the usual definition of the induced norm you describe. $\endgroup$ – VHarisop Nov 4 '18 at 22:19
  • $\begingroup$ 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. $\endgroup$ – michalOut Nov 4 '18 at 22:52
  • $\begingroup$ yes VHarisop, there is a squareroot, I wasnt sure it was relevant. $\endgroup$ – vulcan583 Nov 5 '18 at 0:27
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.