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This question is the follow up question to Operator norm increases under taking absolute value of all entries of a matrix, which (correctly) hypothesizes that for some matrix $A$ if we define matrix $B=(b_{ij})_{ij}$ with $b_{ij}=|a_{ij}|$, then we have $$\|A\| \le \|B\|.$$

Now my question is whether we also have

$$\|A^n\|\le \|B^n\|$$

This is likely true, but I have not found it yet. If I figure it out I will post an answer here.

user1551
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Felix B.
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1 Answers1

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You can use the same trick and conclude in the same way. The components of $A^n v$ are some homogeneous polynomial expressions of the entries of $A$ with the components of $v$ as coefficients, say $$ (A^nv)_i=P_i(a_{11},…,a_{NN};v_1,…,v_N). $$

The components of $B^n u$ (I am using the same notations of the linked question/answers) are of that form as well with the exact same polynomial appearing, i.e., $$ (B^nu)_i=P_i(|a_{11}|,…,|a_{NN}|;|v_1|,…,|v_N|), $$

and you can bound the former ones with the latter ones just by using the triangle inequality.

The explicit expression of $P_i$ is not that hard to write actually: $$ (A^n v)_i=\sum_{1\leq j_1,…,j_n\leq N} a_{ij_1}a_{j_1j_2}…a_{j_{n-1}j_n}v_{j_n}. $$

Lorenzo Pompili
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  • I am also just now realizing that we have more generally $$|\prod_{i=1}^n A^{(i)} v| = \sum_{j=1}^d \Big| \sum_{i_1,\dots,i_n=1,\dots,d} a^{(1)}{j i_1} a^{(2)}{i_1 i_2}\dots a^{(n)}{i{n-1} i_n} v_{i_n} \Big|^2 \le |\prod_{i=1}^n B^{(i)} u| $$ – Felix B. Nov 11 '21 at 17:35
  • You are right! XD Nice – Lorenzo Pompili Nov 11 '21 at 17:40