Skip to main content
9 events
when toggle format what by license comment
Jan 8, 2023 at 17:33 history edited Toni Mhax CC BY-SA 4.0
typos
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
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