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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?

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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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