similar matrices have the same bandwidth?

Question

If $A$ is symmetric with bandwidth $p$ then $A_+ = Q^{T} A Q$, where $Q$ is orthogonal, is orthogonally similar to $A$. How can we show/prove that $A_+$ also has bandwidth $p$ ?

By bandwidth, do you mean this? Also, what is $A_+$? – Ben Grossmann Apr 22 '14 at 23:59 — Ben Grossmann, Apr 22 '14 at 23:59
yes, and I defined $A_+$ as above. – Elnaz Apr 23 '14 at 00:25 — Elnaz, Apr 23 '14 at 00:25
Oh, I see, $Q$ is an arbitrary orthogonal matrix then. – Ben Grossmann Apr 23 '14 at 00:30 — Ben Grossmann, Apr 23 '14 at 00:30

score 2 · Answer 1 · answered Apr 23 '14 at 00:33

2

This is not necessarily the case. For example: the matrices $$ A = \pmatrix{3&0&0\\0&0&0\\0&0&0}; \quad A_+ = \pmatrix{1&1&1\\1&1&1\\1&1&1} $$ Clearly do not have the same bandwidth.

answered Apr 23 '14 at 00:33

Ben Grossmann

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What is the $Q$ here? – Elnaz Apr 23 '14 at 00:46
One valid $Q$ is $$ Q = \pmatrix{ 1/\sqrt 3 & 1 / \sqrt 3 & 1 / \sqrt 3\ 1/\sqrt 2 & -1/\sqrt 2 & 0\ 1/\sqrt 6 & 1/\sqrt 6 & -\sqrt{(2/3)} } $$ To be sure though, any two symmetric matrices with the same eigenvalues are orthogonally similar. – Ben Grossmann Apr 23 '14 at 02:03

similar matrices have the same bandwidth?

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