$P,Q$ are positive definite diagonal matrices. $A,B$ are matrices, then $\lVert PA - QB \rVert_2 \leq \max(\rho(P),\rho(Q)) \lVert A - B \rVert_2$

Question

$P,Q$ are positive definite diagonal $n \times n$ matrices with $\rho(P),\rho(Q) < 1$. $A,B$ are $n \times m$ matrices. Prove that

$\lVert PA - QB \rVert_2 \leq \max(\rho(P),\rho(Q)) \lVert A - B \rVert_2$

where $\lVert \cdot \rVert_2$ is the 2-norm and $\rho$ denotes the spectral radius.

I am not sure if this result is even true - but it would great to get a pointer toward either a proof or a counterexample.

score 1 · Accepted Answer · answered Aug 05 '20 at 17:10

1

This isn't true in general. Try $n = 2$, $A = B = I$, $Q = I$, $P = \pmatrix{2 & 0 \\ 0 & 1}$. Then $PA - QB$ is nonzero, but $A - B$ is zero.

In fact, even for $n = 1$ it seems to be false, for if $a = b = 1$, $p = 2$ and $q = 1$, your formula says that $$ |2-1| < \max(1, 2) |1 - 1|. $$

answered Aug 05 '20 at 17:10

John Hughes

ahh, thank you! – Allen Hart Aug 05 '20 at 17:18
I should add an extra condition then, that the eigenvlaues of $P,Q$ lie in $(0,1)$ – Allen Hart Aug 05 '20 at 17:19
Pick $a = b = 1, p = 0.5, q = 0.6$ and you've still got a counterexample for $n = 1$. I think you should really consider why you believe that this might be true, and keep in mind the case $A = B$ as you examine your reasons. that's always going to make the right-hand side be zero, and it's really tough to prove inequalities with zeroes on the right. – John Hughes Aug 05 '20 at 17:28
actually - even with this extra condition - I think the statement is still false. – Allen Hart Aug 05 '20 at 17:29
ahh you got there before me, thanks again – Allen Hart Aug 05 '20 at 17:29
as you suggest, I will consider this more carefully. – Allen Hart Aug 05 '20 at 17:32

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