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?


1 Answer 1


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 $1$.

  • 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. $\endgroup$
    – Nik Weaver
    Jan 21, 2019 at 19:01

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