Positive definite order and matrix norm induced by weighted vector norm

Question

Let $W$ be an $n$ by $n$ nonsingular matrix that defines a weighted two norm for vectors in $\Re^n$. Define an induced matrix norm by this weighted norm. Suppose $A$ and $B$ are positive semi-definite matrices such that $A\geq B$ ($A-B$ is positive semidefinite), is it true that $\Vert A\Vert \geq \Vert B\Vert$? I know for the case where $W$ is the identity matrix, the inequality would hold, but cannot show the inequality for the weighted case.

Answer 1

Denote $\|x\|_W = \|Wx\|_2$. The induced norm can be described as $$ \|A\| = \max_{x \neq 0} \frac{\|Ax\|_W}{\|x\|_W} = \max_{x \neq 0}\frac{\|WAx\|_2}{\|Wx\|_2} = \max_{y \neq 0} \frac{\|WAW^{-1}y\|_2}{\|y\|_2} = \|WAW^{-1}\|_2. $$ With that in mind, consider $$ A = \pmatrix{3&0\\0&3}, \quad B = \pmatrix{1&1\\1&1}, \quad W_t = \pmatrix{t&0\\0&1}. $$ Note that $A \geq B$ (and indeed, $A > B$). On the other hand, note that for any matrix $M$, $\|M\|_2 \geq |m_{ij}|$ for all entries $m_{ij}$ of $M$. Thus, we find that the matrix $$ M = W_tBW_t^{-1} = \pmatrix{1 & t\\t^{-1} & 1} $$ satisfies $\|M\| \geq t$ for $t > 0$. Thus, with $t = 4$ for example, we find that $$ \|A\|_{W_t} = \|W_tAW_t^{-1}\|_2 = \|A\|_2 = 3 < t \leq \|B\|_{W_t}. $$