Let $A$ e $B$ be two symmetric matrices, with the additional property that the sum of all the entries of any column is zero.

If the $k-$th column of $A$ equals the $k-$th column of $B$, what can be said about the $k-$th columns of $\exp(A)$ and $\exp(B)$?

From some tinkering in Mathematica, it looks like the equality should be preserved even after the exponentiation. If that is indeed the case, how does one go about proving it?


This is false. For instance, let $$A=\begin{pmatrix} 1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}$$ and $$B=\begin{pmatrix} 1 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & -1\end{pmatrix}.$$ These have the same first column but $\exp(A)$ and $\exp(B)$ do not have the same first column.

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  • $\begingroup$ Thanks for replying. I omitted to mention a few important additional facts. The matrices I'm interested in are Laplacian matrices of simple, connected, finite graphs. So, for example, $A$ doesn't work because it has an all-zeroes column. I should probably edit the question accordingly; I thought things could be a little more general than that... $\endgroup$ – TotalNoob Jul 11 at 13:10

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