2
$\begingroup$

I am working with a product of $n\times n$ matrices $A_1,\ldots,A_k$. Under which conditions can I assume that

$$\|A_1\cdots A_k\|_\infty \leq \|A_1\cdots \hat{A_i}\cdots A_k\|_\infty \|A_i\|_\infty,$$

where $\|\cdot\|_\infty$ denotes the operator norm, and $\hat{A_i}$ denotes the omission of $A_i$.

E.g. if all products of subsets of $\{A_1,\ldots,A_k\}$ are normal, then the above inequality should follow from the submultiplicativity of the operator norm and the fact that $AB$ and $BA$ have the same singular values for normal $A$ and $B$. However, this condition seems rather artifical and I am wondering if something more powerful holds. What if all $A_i$ are normal, self-adjoint, unitary or orthogonal projectors?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

For a simple counterexample, suppose $A_1$, $A_2$, $A_3$ are the orthogonal projections $$ A_1 = \pmatrix{1 & 0\cr 0 & 0\cr},\ A_2 = \pmatrix{1/2 & 1/2\cr 1/2 & 1/2\cr},\ A_3 = \pmatrix{0 & 0\cr 0 & 1\cr}$$ Then $\|A_1 A_3\| = 0$ but $\|A_1 A_2 A_3\| > 0$.

That covers all your cases except "unitary". Of course if the $A_i$ are all unitary, both sides of your inequality are $1$.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ A famous example from quantum mechanics! $A_1$ is vertical polarization, $A_2$ is diagonal, and $A_3$ is horizontal. $A_1$ followed by $A_3$ blocks everything, but if you interpose $A_2$ then something gets through. $\endgroup$ Commented Jan 21, 2019 at 19:01

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.