# Is it possible to show that exp(A)-exp(B) is negative definite provided $A-B$ is negative definite? [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, John B, Peter, Chris Custer, B. Mehta
If this question can be reworded to fit the rules in the help center, please edit the question.

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.
• 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. – Shalop Jan 26 '18 at 22:48
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.