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?



Bound for psd-matrix weighted norm of two related vectors
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 $x'\mathbf{M}x = 20 > \lambda^2 y'\mathbf{M}y = 17$.


