3

If $A$ is a psd matrix, can we say $F(A,T) :=\int_0^T e^{tA}\,dt = (e^{TA}-I)A^\dagger$, where $A^\dagger$ is the Moore-Penrose pseudo-inverse of $A$ ?

Afterall,

  • The result holds if $A$ is positive-definite (thanks to the first part of this post https://math.stackexchange.com/a/658289/168758).
  • For small $\lambda>0$ can approximate $A$ by a positive-definite matrix $B_\lambda := A+\lambda I$, and it argue that $$ F(A,t) \overset{?}{=} \lim_{\lambda \to 0^+}F(B_\lambda, T) = \lim_{\lambda \to 0^+}(e^{TB_\lambda}-I)B_\lambda^{-1} = (e^{TA}-I)A^\dagger, $$ where the first step is not justified.
dohmatob
  • 9,753

2 Answers2

5

The notations $\frac{e^{At}-}{A}$ or $(e^{At} - )A^{-1}$ are typically just used as a (bad) way to abbreviate the series

$$\begin{aligned} \frac{ + At + ½A^2t^2 + … -}{A} = \frac{At(+½At + …)}{A} = t\sum_{k=0}^{∞}\tfrac{(At)^k}{(k+1)!} \end{aligned}$$

which exists even when $A$ is arbitrarily singular. This quantity occurs when solving linear ODEs.

A better notation would be to use the so-called $φ$ functions, which are recursively defined by

$$ φ_m(z) = zφ_{m+1}(z) + \tfrac{1}{m!}，φ_0(z) = e^z \quad⇝\quadφ_m(z) = \sum_{k=0}^∞ \frac{z^k}{(k+m)!}$$

For example, the solution to the inhomogeneous linear ODE with constant coefficients

$$ \dot{x}(t) = Ax(t) + b$$

Is given by $x(t+∆t) = e^{A{∆t}}x(t) + φ(A{∆t}){∆t}b$.

More generally, if $b=b(t)$ is time-varying and locally analytical on $U_{∆t}(t)$, then the solution to $\dot{x}(t) = Ax(t) + b(t)$ in this neighbourhood can be expressed as

$$ x(t+∆t) = e^{A{∆t}}x(t) + \sum_{k=1}^{∞}φ_k(A{∆t}) {∆t}^k \frac{^{k-1} b(t)}{ t^{k-1}}$$

References:

Hyperplane
  • 12,204
  • 1
  • 22
  • 52
  • Interesting (as indeed my problem stems from solving linear ODEs). Thanks. So, when the dust has settled, are you saying that $\int_0^T e^{tA},ds = t\sum_{k=0}^\infty \frac{(At)^k}{(k+1)!}$, irrespectively of whether $A$ is singular or not ? – dohmatob Jun 01 '22 at 18:49
  • 1
    @dohmatob See my edit. – Hyperplane Jun 01 '22 at 19:01
  • OK, thanks for the very effective refresher on ODEs. This solves my problem! – dohmatob Jun 01 '22 at 19:04
  • @dohmatob Regarding $\int_0^T e^{tA},ds = t\sum_{k=0}^\infty \frac{(At)^k}{(k+1)!}$, don't be afraid to check whether we are allowed to integrate a series term-by-term. – Hyperplane Jun 01 '22 at 19:08
  • Indeed, $A$ is finite-dimensional operator therefore compact, and so $|(At)^k| = |A| t^k$, with $|A| < \infty$. – dohmatob Jun 01 '22 at 19:11
  • 1
    $+\tt1$ Great answer. The interesting question is why multiplying $(e^A-I)$ by the pseudoinverse does not equal $\varphi(A)$ when $A$ is singular. – greg Dec 19 '23 at 19:46
3

No, it is not true. By a change of orthonormal basis, we may assume that $A$ is a nonnegative diagonal matrix. The problem therefore reduces to the scalar case.

When $a>0$, straightforward calculation shows that $\int_0^Te^{ta}dt=(e^{Ta}-1)a^{-1}$. However, when $a=0$, we have $\int_0^Te^{ta}dt=T\ne0=(e^{ta}-1)a^+$. Therefore the equality in question holds only if $A$ is positive definite. When $A$ is positive semidefinite but singular, the correct identity is $$ \int_0^Te^{tA}dt=(e^{TA}-1)A^++T(I-A^+A). $$

user1551
  • 149,263