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Let two vectors, $\mathbf{x}, \mathbf{y}$ be related as: $0 \leq x_i \leq \lambda y_i$, for some $\lambda > 0$. That is, $\mathbf{x}$ is coodinate-wise dominated by a scaled version of $\mathbf{y}$.

Also let $\mathbf{M}$ be a positive semi-definite matrix. My question is, how are their matrix $\mathbf{M}$ induced weighted norms related? I think they are related as: $x\mathbf{M}x' \leq \lambda^2 y\mathbf{M}y'$. Yet I am so far unable to prove this. How do I prove (if above relation is not correct, then a correct version) this relation?

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  • $\begingroup$ Does my answer, in its current form (counterexample with all nonngative elements), adequately address your question? $\endgroup$ Commented Jun 26, 2018 at 21:29

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This is trivially false even in one dimension. Let $M = [1], \lambda = 1$. Then $x = -2, y = 1$ provides a counterexample.

Now let's change the condition to $|x_i| \le |\lambda y_i|$ coordinate-wise. Then $x'\mathbf{M}x \leq \lambda^2 y'\mathbf{M}y$ is true if $M$ is diagonal positive semidefinite (I changed to using column vectors). However, this result is still not ture for a general non-diagonal symmetric positive semidefinite $M$. Let $M =\begin{bmatrix}2 & 1\\1 & 1\end{bmatrix}$, $x = \begin{bmatrix}-1\\0\end{bmatrix}$, $y = \begin{bmatrix}-1\\1\end{bmatrix},\lambda = 1$. Then $x'\mathbf{M}x = 2 > \lambda^2 y'\mathbf{M}y = 1$.

Edit: I am adding another counterexample given the new condition $0 \le x$.

$Edit^2$: Swapped out previous counterexample for one with all integer entries.

Let $M =\begin{bmatrix}1 & -1\\-1 & 2\end{bmatrix}$, $x = \begin{bmatrix}2\\4\end{bmatrix}$, $y = \begin{bmatrix}5\\4\end{bmatrix},\lambda = 1$. Then $x'\mathbf{M}x = 20 > \lambda^2 y'\mathbf{M}y = 17$.

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  • $\begingroup$ Thanks for the answer. I forgot to write that in my case, $x_i, y_i$ are all non-negative. I have edited my question to fix this. I am interested in the general case (not just diagonal matrices). Your counter example for a general psd matrix $\mathbf{M}$ is informative. So then, are the various weighted norms equivalent to the usual norm? That is, are there positive constants $c_1, c_2$, so that $c_1 xx' \leq x\mathbf{M}x' \leq c_2 xx', \forall{x}$? And from that can we derive an upper bound for $x\mathbf{M}x'$ in terms of $y\mathbf{M}y'$ for any psd matrix $M$? $\endgroup$
    – user125930
    Commented Jun 24, 2018 at 18:06
  • $\begingroup$ In my case, $0 \leq x_i \leq \lambda y_i$ $\endgroup$
    – user125930
    Commented Jun 24, 2018 at 18:19
  • $\begingroup$ In my case, $0 \leq x_i \leq \lambda y_i$, so does the condition hold that $x\mathbf{M}x' \leq \lambda^2 y\mathbf{M}y'$?. Your counter example was for a general $x, y$ with negative entries. $\endgroup$
    – user125930
    Commented Jun 24, 2018 at 18:26

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