Timeline for On the bounds of the sum of the squares of spectral variation of two real symmetric matrices
Current License: CC BY-SA 4.0
9 events
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Jan 8, 2023 at 17:33 | history | edited | Toni Mhax | CC BY-SA 4.0 |
typos
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Jan 6, 2023 at 0:56 | comment | added | Brendan McKay | Except for $B=0$, yes. Say $B$ is a counterexample and replace it by $-B$ if necessary so that $\lambda_1$ has the greatest absolute value. For integer $k$, the trace of $B^{2k}$ is $\sum_i{\lambda_i^{2k}}$. If all the eigenvalues are less than 1 in absolute value, and not all zero, this expression will be strictly between 0 and 1 for large enough $k$. But the trace of $B^{2k}$ is an integer, so there is no such $B$. | |
Jan 5, 2023 at 16:05 | comment | added | shahulhameed | @BrendanMcKay I mean the largest eigenvalue of any $B$ will be at least one. i.e., $\lambda_1(B)\ge 1$ | |
Jan 5, 2023 at 10:14 | comment | added | Brendan McKay | @shahulhameed The eigenvalues can range from $-n+1$ to $n-1$, including 0. | |
Jan 5, 2023 at 9:19 | comment | added | shahulhameed | @Toni Mhax If we can prove that maximum eigenvalue of these matrices have least upper bound 1, then I have the proof for the bound. Can anybody help me | |
Jan 4, 2023 at 7:32 | comment | added | shahulhameed | @Brendan Mckay Yes, I strongly feel that $2n(n-2)$ is a tight bound which is attained for the matrices $A=J-I$ and $B=-A$. The question is whether we can prove it by some means; I hope matrix norms would help us by intuition | |
Jan 4, 2023 at 6:36 | history | edited | Toni Mhax | CC BY-SA 4.0 |
added 46 characters in body
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Jan 4, 2023 at 6:27 | comment | added | Brendan McKay | This is a correct proof of the bound $4n(n-1)$ but it is not attained by $A=J-I, B=-A$. You need to sort the eigenvalues separately for $A$ and $B$. For $A$ it is $n-1,-1,\ldots,-1$ and for $B$ it is $1,\ldots,1,-(n-1)$. This gives $2n(n-2)$ as OP said. | |
Jan 4, 2023 at 5:41 | history | answered | Toni Mhax | CC BY-SA 4.0 |