= Solution
An <eigenvalue> $\lambda$ and associated <eigenvector> $e\ne0$ satisfy $Ae=\lambda e$. The <characteristic polynomial> $\det(A-\lambda I)$ has degree equal to the positive <matrix> size. The <fundamental theorem of algebra> gives a complex root, at which $A-\lambda I$ is singular and therefore has a nontrivial kernel. Hence every nonempty square complex <matrix> has an <eigenvector>.
A nondiagonalizable example is the <Jordan block> $\begin{pmatrix}0&1\\0&0\end{pmatrix}$: its only <eigenvalue> is zero, and its eigenspace is one-dimensional. A <Hermitian matrix> satisfies $A^\dagger=A$ and has an <orthonormal eigenbasis> by the <spectral theorem>. One proof finds an <eigenvector>, notes that its <orthogonal complement> is invariant, and repeats. Its <eigenvalues> are real because
$$
\lambda\,e^\dagger e=e^\dagger Ae=\overline{e^\dagger Ae}=\overline\lambda\,e^\dagger e,
$$
where $e^\dagger e>0$.
With all <eigenvalues> negative, $A$ is invertible. Its same complete eigenbasis gives
$$
\boxed{A^{-1}e_i=\lambda_i^{-1}e_i,\qquad A^me_i=\lambda_i^me_i\quad(m=1,2,\ldots).}
$$
Some powered <eigenvalues> can coincide, but the original <eigenvectors> remain a complete basis. Writing $A=U\operatorname{diag}(\lambda_i)U^\dagger$, the convergent <matrix exponential> series gives
$$
\int_0^\infty e^{tA}dt=U\operatorname{diag}\left(\int_0^\infty e^{t\lambda_i}dt\right)U^\dagger=U\operatorname{diag}(-\lambda_i^{-1})U^\dagger.
$$
Thus the <matrix inverse from a decaying matrix exponential> is
$$
\boxed{\int_0^\infty e^{tA}dt=-A^{-1},\qquad A^{-1}=-\int_0^\infty e^{tA}dt.}
$$
\b[The identity printed in the original PDF is missing a minus sign.] Already for the one-by-one <matrix> $A=(-1)$ its integral is $1$, whereas its inverse is $-1$. Equivalently, integrating $d(e^{tA})/dt=Ae^{tA}$ gives $A\int_0^\infty e^{tA}dt=-I$, confirming the correction without diagonalization.
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