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Let $T\in M_{n}(\mathbb{R})$ be a tridiagonal matrix. What can we say about operator norm $\|T\|_2$?

I'm asking this question because we know that if $T$ were only diagonal, then $\|T\|_2$ is the largest absolute value of any diagonal entry of $T$, as shown here, and so there might also a similar result for tridiagonal or $k$-diagonal matrices in general.

UserA
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  • For a general matrix $A\in M_n(\mathbb{R})$, the value $||A||_2$ is equal to the square root of the largest eigenvalue of $A^TA$ over $\mathbb{C}$. (all the eigenvalues of $A^TA$ are non negative real numbers, so we can compare them). Not sure this can be described by an easy formula for a tridiagonal matrix. – Mark Jul 03 '20 at 12:48
  • @RodrigodeAzevedo No any tridiagonal matrix in general. – UserA Jul 03 '20 at 12:55
  • @Mark what about an upper bound? suppose that I have a uniform upper bound $M$ for all entries. Can I find an upper bound for $|T|_2$ using $M$ only and independent of $n$? – UserA Jul 03 '20 at 12:58
  • Sorry, what is a k-diagonal? – chhro Jul 03 '20 at 19:00
  • @chhro tridiagonal is a matrix with 3 diagonals, pentadiagonal with 5 and $k$-diagonal with $k$ diagonals. – UserA Jul 04 '20 at 05:49

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Edit. As user8675309 rightly points out, using Schur's test $\|A\|_2\le\sqrt{\|A\|_1\|A\|_\infty}$, we get $\|A\|_2\le3\max_{i,j}|a_{ij}|$. This bound is at least asymptotically tight: let $A_n$ be the $n\times n$ symmetric tridiagonal Toeplitz matrix with all entries on the three diagonals equal to $1$. Then $\lim_{n\to\infty}\|A_n\|_2=3$.

For a $(2k+1)$-diagonal matrix, Schur's test gives $\|A\|_2\le\min(n,\,2k+1)\max_{i,j}|a_{ij}|$.

(Old answer) Here is an obvious upper bound: since $A$ is tridiagonal, when we calculate $(AA^T)_{ij}$ as the sum $\sum_ka_{ik}a_{jk}$, at most three summands are involved. Also, as $AA^T$ is pentadiagonal, if we apply Gerschgorin disc theorem and the triangle inequality, we get $\rho(A^TA)\le(5)(3)\max_{i,j}|a_{ij}|^2$. Consequently, $$ \|A\|_2=\sqrt{\rho(AA^T)}\le\sqrt{15}\max_{i,j}|a_{ij}|. $$

user1551
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  • @user15551 Thank you for the answer! Can this bound be generalized to $2k+1$-diagonal matrices? For example, I've encountered a 7-diagonal matrix. In this case, how many diagonals does $AA^T$ have? – UserA Jul 03 '20 at 16:18
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    @UserA By an analogous argument, $|A|2\le\sqrt{(2k+1)\min(n,4k+1)}\max{i,j}|a_{ij}|$. – user1551 Jul 03 '20 at 16:27
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    note: this upper bound can be tightened from $\sqrt{15}\max_{i,j}|a_{ij}\vert$ to $3\max_{i,j}|a_{ij}\vert$ if one instead uses the Schur Test. – user8675309 Jul 03 '20 at 18:21