Given a symmetric $N\times N$ matrix $A$, with eigenvalues $-x_1,-x_2,-x_3,\dots,-x_N$ and $x_1,x_2,\dots,x_N >0$.
$A$ is known as a negative-definite matrix.
We can diagonalize $A$ as $A = W*D*W^{-1}$
$D$ is a diagonal matrix with $A$'s eigenvalues on the diagonal position and all other elements are $0$. $W$ is composed by $A$'s eigenvector corresponding to eigenvalues of $A$.
how can I prove that $(I-e^{At})^{-1}$ only contains positive elements? if $I$ is a $N\times N$ identity matrix, $t>0$.
The current progress I have is
$$e^{At} = W* (e^{Dt}) *W^{-1}$$
$$I-e^{At} = W* (e^{I-Dt}) *W^{-1}$$
$e^{At}$ only contains positive elements
How to prove that $(I-e^{At})^{-1}$ only contains positive elements? Thank you
For $t>0$ we can use the geometric series formula:
$$(I-e^{At})^{-1}=\sum_{n=0}^\infty \left ( e^{At} \right )^n.$$
where by convention $B^0=I$ for a square matrix $B$. Here we have used that $\| e^{At} \| < 1$ for $t>0$ in the Euclidean operator norm. This follows because the eigenvalues of $e^{At}$ are $e^{\lambda t}$ for eigenvalues $\lambda$ of $A$, and because the Euclidean operator norm is the largest singular value, which for a symmetric matrix is the eigenvalue of largest magnitude.
So it follows from what you've already shown (that $e^{At}$ has only positive elements).
The geometric series formula for matrices is proven the same way as the geometric series formula for numbers: prove that $\sum_{n=0}^N A^n = (I-A^{N+1})(I-A)^{-1}$ and then send $N \to \infty$.