Inverse of Hermitian matrix with all negative eigenvalues?

Question

I am asked to prove that for a Hermitian matrix with all eigenvalues negative, the inverse is given by

$$A^{-1} = - \int_0^\infty e^{tA} {\text { d}}t.$$

(I have the corrected the missing minus sign above)

I have tried expanding out the integrand in series form, and can show that without taking into account the limits, I have $e^{tA}-I$ as the antiderivative, but I don’t really know how to play around with the limits to get a non-divergent solution. I was thinking of transforming to the diagonal basis, but I can’t seem to move on from there. Is there a trick I’m missing out on? Thanks!

Answer 1

There is more than one way to do this. I like to exploit $$Ae^{tA} = \frac{d}{dt} e^{tA}.$$ Hence $$ A \int_0^T e^{tA}dt = \int_0^T \frac{d}{dt} e^{tA} dt = [ e^{tA}]_0^T = e^{TA} - I \rightarrow - I, \quad T \rightarrow \infty, \quad T \in \mathbb{R},$$ provided the eigenvalues of $A$ are strictly negative. It follows that $$ A^{-1} = - \int_0^\infty e^{tA}dt.$$

As pointed out by @JoséCarlosSantos there is a misprint in your problem. The intended statement may well have been $$ A^{-1} = \int_0^M e^{-tA}dt.$$

Answer 2

The result seems to hold for any matrix with all negative eigenvalues.

We have $$- A \int_0^\infty e^{tA}\,dt = - \int_0^\infty Ae^{tA}\,dt = -e^{tA} \Bigg|^\infty_0 = e^0 - \lim_{t\to\infty} e^{tA} = I - \lim_{t\to\infty} e^{tA}$$

If $A$ has an $n\times n$ Jordan block $$J =\pmatrix{ \lambda & 1 & 0 & \cdots & 0 & 0\\ 0 & \lambda & 1 & \cdots & 0 & 0\\ 0 & 0 & \lambda &\cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 &\cdots & \lambda & 1\\ 0 & 0 & 0 &\cdots & 0 & \lambda\\ }$$

then $$e^{tJ} = \pmatrix{ e^{t\lambda} & te^{t\lambda} & \frac{t^2}2e^{t\lambda} & \cdots & \frac{t^{n-2}}{(n-2)!}e^{t\lambda} & \frac{t^{n-1}}{(n-1)!}e^{t\lambda}\\ 0 & e^{t\lambda} & te^{t\lambda} & \cdots & \frac{t^{n-3}}{(n-3)!}e^{t\lambda} & \frac{t^{n-2}}{(n-2)!}e^{t\lambda}\\ 0 & 0 & e^{t\lambda} &\cdots & \frac{t^{n-4}}{(n-4)!}e^{t\lambda} & \frac{t^{n-3}}{(n-3)!}e^{t\lambda}\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 &\cdots & e^{t\lambda} & te^{t\lambda}\\ 0 & 0 & 0 &\cdots & 0 & e^{t\lambda}\\ }$$

so since $\lambda < 0$ clearly $$\lim_{t\to\infty} e^{tJ} = 0$$

because the exponential decays faster than any polynomial. Hence $\lim_{t\to\infty} e^{tA} = 0$.

Therefore $$- A \int_0^\infty e^{tA}\,dt = I \implies A^{-1} = -\int_0^\infty e^{tA}\,dt$$

Answer 3

For nonsingular matrices $A$, the antiderivative of $e^{At}$ is not $e^{At} - I$ but rather $A^{-1}e^{At}$:

$(A^{-1}e^{At})' = A^{-1}(e^{At})' = A^{-1}Ae^{At} = e^{At}; \tag 1$

then we have

$\displaystyle \int_0^\infty e^{At}\; dt = \lim_{\tau \to \infty} \int_0^\tau e^{At} \; dt = \lim_{\tau \to \infty} \int_0^\tau (A^{-1}e^{At})' \; dt$ $= \displaystyle \lim_{\tau \to \infty}(A^{-1}e^{At} \mid_0^\tau = \lim_{\tau \to \infty} (A^{-1}e^{A\tau} - A^{-1} e^0) = A^{-1} \lim_{\tau \to \infty} (e^{A\tau} - I). \tag{2}$

Now since $A$ is Hermitian, it may be diagonalized via a unitary matrix $U$, and the eigenvalues $\mu_i$ of $A$ will form the diagonal of the resulting diagonal matrix $D$, that is

$D = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n) = UAU^\dagger, \tag 3$

whence

$A = U^\dagger D U = U^\dagger \text{diag}(\mu_1, \mu_2, \ldots, \mu_n) U,\tag 4$

where every $\mu_i < 0$, $1 \le i \le n$; then for any $t \in \Bbb R$ we have

$At = U^\dagger Dt U = U^\dagger \text{diag}(\mu_1 t, \mu_2 t, \ldots, \mu_n t) U,\tag 5$

$(At)^k = U^\dagger (Dt)^k U = U^\dagger \text{diag}((\mu_1 t)^k, (\mu_2 t)^k, \ldots, (\mu_n t)^k) U,\tag 6$

$\dfrac{(At)^k}{k!} = U^\dagger \dfrac{(Dt)^k}{k!} U = U^\dagger \text{diag}\left ( \dfrac{(\mu_1 t)^k}{k!}, \dfrac{(\mu_2 t)^k}{k!}, \ldots, \dfrac{(\mu_n t)^k}{k!} \right) U,\tag 7$

and finally,

$e^{At} = \displaystyle \sum_0^\infty \dfrac{(At)^k}{k!} = U^\dagger \left ( \sum_0^\infty \dfrac{(Dt)^k}{k!} \right )U$ $= U^\dagger \text{diag}\left ( \displaystyle \sum_0^\infty \dfrac{(\mu_1 t)^k}{k!}, \sum_0^\infty \dfrac{(\mu_2 t)^k}{k!}, \ldots, \sum_0^\infty \dfrac{(\mu_n t)^k}{k!} \right)U = U^\dagger (e^{\mu_1 t}, e^{\mu_2 t}, \ldots, e^{\mu_n t})U; \tag 8$

now since every $\mu_i < 0$, we find that

$\displaystyle \lim_{t \to \infty} e^{At} = \lim_{t \to \infty} U^\dagger (e^{\mu_1 t}, e^{\mu_2 t}, \ldots, e^{\mu_n t})U \to 0. \tag 9$

We may now deploy (9) in (2):

$\displaystyle \int_0^\infty e^{At}\; dt = A^{-1} \lim_{\tau \to \infty} (e^{A\tau} - I) = A^{-1}(-I) = -A^{-1}, \tag{10}$

as per request.