Is $e^{-(L+L^T)}$ a positive definite matrix?

Question

$L$ is a nonsymmetric Laplacian matrix and the zero eigenvalue is simple, is it true that $e^{-(L+L^T)}$ a positive definite matrix (I can be sure that it is non-negative matrix)?

Answer 1

Let $S=L+L^T$, then $S$ is symmetric, hence all eigenvalues of $S$ are real.

It should be clear, that $e^{-S}$ is symmetric. Now let $ \lambda$ an eigenvalue of $e^{-S}$. Then there is an eigenvalue $ \mu$ of $S$ such that $\lambda = e^{-\mu}$.

Since $ \mu$ is real, we have that $ \lambda >0$.

Conclusion: $e^{-S}$ is positive definite.

Answer 2

If $A$ is a real symmetric matrix, then $\exp(A)$ is positive definite. For $A=UDU^{-1}$ where $D$ is diagonal and $U$ is orthogonal, and then $\exp(A)=U\exp(D)U^{-1}$.

Answer 3

Here is a proof for real matrices $L$, whether or not the $0$ eigevalue is simple:

Note that, for any real matrix $L$, $L + L^T$ is symmetric, hence is diagonalized by some orthogonal matrix $O$:

$O^T(L + L^T)O = \Lambda, \tag 1$

where

$\Lambda = [\lambda_i \delta_{ij}]; \tag 2$

here the $\lambda_i$ are the eigenvalues of $L + L^T$; since $L + L^T$ is symmetric, $\lambda_i \in \Bbb R$ for all values of the index $i$; then

$O^T e^{-(L + L^T)} O = e^{-\Lambda}; \tag 3$

notice that

$e^{-\Lambda} = [e^{-\lambda_i} \delta_{ij}]; \tag 4$

furthermore,

$e^{-\lambda_i} > 0, \; \forall \lambda_i; \tag 5$

we thus see that $e^{-\Lambda}$ is a positive definite diagonal matrix. From (3),

$e^{-(L + L^T)} = O e^{-\Lambda}O^T; \tag 6$

it thus follows that, for any vector $y$,

$\langle y, e^{-(L + L^T)}y \rangle = \langle y, Oe^{-\Lambda}O^Ty \rangle = \langle O^Ty, e^{-\Lambda}O^Ty \rangle > 0, \tag 7$

which says that $e^{-(L + L^T)}$ is positive definite.