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If two symmetric square matrices $A$ and $B$ are such that $A-B$ is negative definite, I am to prove that $\exp(A)-\exp(B)$ is negative definite. I initially used the following exponential series

$e^{A}=\sum_{i\geq{0}}\dfrac{A^{i}}{i!}$ and $e^{B}=\sum_{i\geq{0}}\dfrac{B^{i}}{i!}$

to prove the above claim. But this approach did not help. Any hint on the approach will be much appreciated.

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closed as off-topic by Namaste, John B, Peter, Chris Custer, B. Mehta May 5 '18 at 0:06

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The example given in this answer serves as a counterexample. In particular: if we take $$ A = \pmatrix{1 & 0\\0 & 2}, \quad B = \pmatrix{2 + \epsilon & 1\\1 & 3} $$ (e.g. with $\epsilon = 0.01$), we find that $A - B$ is negative definite, but $\exp(A) - \exp(B)$ fails to be negative definite.

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    $\begingroup$ This is not relevant to your answer, but if $A,B$ are symmetric and they commute, then the claim is true, no? By simultaneous diagonalization. $\endgroup$ – Shalop Jan 26 '18 at 22:48
  • $\begingroup$ @Shalop that's right $\endgroup$ – Omnomnomnom Jan 26 '18 at 23:01
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Thanks for the response. If A and B are symmetric, they are diagonalizable and if they commute, they are simultaneously diagonalizable. This implies there exists a nonsingular square matrix P such that $D_{A}$=$P^{-1}AP$ and $D_{B}=P^{-1}BP$ are diagonal with the eigenvalues of A and B respectively in their diagonal elements.

It is easy to verify that $\lambda_{i}(D_{A})<\lambda_{i}(D_{B})$ due to the negative definiteness property of A-B, where $\lambda_{i}$ is the $i^{\text{th}}$ eigenvalue. As $exp(\lambda_{i}(D_A))<exp(\lambda_{i}(D_B))$, it implies $exp(A)-exp(B)$ is negative definite. I am trying to make sure if the proof goes like this. Thanks again.

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