Last time we derived a formula for the derivative of the matrix exponential. Here we will be focusing instead on the expression $$D\exp(x)u=\exp(x)u=u\exp(x),$$ which holds whenever \(u\) commutes with \(x\). In this post, \(E\) denotes a real Banach space and \(L(E)\) denotes the space of continuous linear operators on \(E\).

## ODEs

The (matrix) exponential can be used to solve certain types of first-order linear systems of ordinary differential equations with non-constant coefficients: not only can we solve \begin{align}x'(t)&=x(t)+y(t) \\ y'(t)&=x(t)+y(t),\end{align} but we can also solve \begin{align}x'(t)&=tx(t)+y(t) \\ y'(t)&=x(t)+ty(t).\end{align}

**Theorem 1.** *Let \(I\) be a connected open subset of \(\mathbb{R}\), let \(f:I\to E\) be differentiable and suppose that \(f'(t)=A(t)f(t)+b(t)\) for all \(t\in I\), where \(A:I\to L(E)\) and \(b:I\to E\) are continuous. Assume that \(A(s)A(t)=A(t)A(s)\) for all \(s,t\in I\). Choose any \(a\in I\). Then there exists some \(c\in E\) such that $$
f(t)=e^{\widehat{A}(t)} \left( c+\int_a^t e^{-\widehat{A}(s)}b(s)\,ds \right),
$$ where $$
\widehat{A}(t)=\int_a^t A(s)\,ds.
$$*

*Proof.* Choose any \(a\in I\) and let $$

g(t)=e^{-\widehat{A}(t)}f(t)-\int_a^t e^{-\widehat{A}(s)}b(s)\,ds.

$$ It is easy to verify that \(A(t)=\widehat{A}'(t)\) commutes with \(\widehat{A}(t)\) for every \(t\in I\), so \begin{align}

g'(t) &= e^{-\widehat{A}(t)}(-\widehat{A}'(t))f(t)+e^{-\widehat{A}(t)}f'(t)-e^{-\widehat{A}(t)}b(t) \\

&= -e^{-\widehat{A}(t)}A(t)f(t)+e^{-\widehat{A}(t)}(A(t)f(t)+b(t))-e^{-\widehat{A}(t)}b(t) \\

&= 0

\end{align} and \(g\) is constant (see this theorem). \(\square\)

Suppose we have a (non-homogeneous) linear system of ODEs with *constant* coefficients. Then \(A(t)\) is constant and \(\widehat{A}(t)=tA\), so $$

f(t)=e^{tA} \left( c+\int_a^t e^{-sA}b(s)\,ds \right).

$$ If the system is homogeneous, then \(b=0\) and we simply have $$

f(t)=e^{tA}c.

$$ On the other hand, if \(E=\mathbb{R}\) then we just have the “integrating factor” method for solving ODEs of the form $$

x'(t)+P(t)x(t)=Q(t).

$$

**Example.** Consider the system \begin{align}

x'(t) &= 2tx(t)+y(t) \\

y'(t) &= y(t)+2tx(t),

\end{align} which can be written as $$

\begin{bmatrix}x'(t) \\ y'(t)\end{bmatrix} = \begin{bmatrix}2t & 1 \\ 1 & 2t\end{bmatrix}\begin{bmatrix}x(t) \\ y(t)\end{bmatrix} = A(t) \begin{bmatrix}x(t) \\ y(t)\end{bmatrix}.

$$ It is easy to see that \(A(t)A(s)=A(s)A(t)\) for all \(s,t\in\mathbb{R}\). We have $$\widehat{A}(t) = \begin{bmatrix}t^2 & t \\ t & t^2\end{bmatrix}.$$ Therefore the general solution is \begin{align}

\begin{bmatrix}x(t) \\ y(t)\end{bmatrix} &= e^{\widehat{A}(t)}\begin{bmatrix}C_1 \\ C_2\end{bmatrix} \\

&= \begin{bmatrix}

\frac{1}{2}C_{1}(e^{(t+1)t}+e^{(t-1)t})+\frac{1}{2}C_{2}(e^{(t+1)t}-e^{(t-1)t}) \\

\frac{1}{2}C_{1}(e^{(t+1)t}-e^{(t-1)t})+\frac{1}{2}C_{2}(e^{(t+1)t}+e^{(t-1)t})

\end{bmatrix}.

\end{align}

## Determinant of the exponential

We now assume that \(V\) is a finite-dimensional real vector space.

**Lemma 1.** *Let \(U\) be the open set of invertible operators in \(L(V)\). Then \(\det:U\to\mathbb{R}\) is differentiable, and $$
D\det(\tau)u=\det(\tau)\operatorname{tr}(\tau^{-1}u).
$$*

*Proof.* Let \(\iota\) be the identity map on \(V\). It is easy to see that \(\det\) is differentiable, by choosing a basis for \(V\). Let \(f(s)=\det(\tau+su)=\det(\tau)\det(\iota+s\tau^{-1}u)\). Then $$

D\det(\tau)u=f'(0)=\det(\tau)\operatorname{tr}(\tau^{-1}u)

$$ since \(\det(\iota+s\tau^{-1}u)\) is a polynomial in \(s\) where the coefficient of \(s\) is \(\operatorname{tr}(\tau^{-1}u)\). \(\square\)

As a simple application of Theorem 1, we prove a well-known formula:

**Theorem 2.** *For all \(\tau\in L(E)\), we have $$
\det(\exp(\tau))=\exp(\operatorname{tr}(\tau)).
$$*

*Proof.* The exponential function is a map \(\exp:L(V)\to U\). Define \(\gamma:\mathbb{R}\to L(V)\) by \(\gamma(s)=s\tau\). Then \begin{align}

(\det\circ\exp\circ\gamma)'(s) &= D\det(\exp(s\tau))(D\exp(s\tau)\tau) \\

&= \det(\exp(s\tau))\operatorname{tr}(\exp(s\tau)^{-1}\exp(s\tau)\tau) \\

&= (\det\circ\exp\circ\gamma)(s)\operatorname{tr}(\tau).

\end{align} Therefore $$

(\det\circ\exp\circ\gamma)(s)=\exp(s\operatorname{tr}(\tau))

$$ by Theorem 1. \(\square\)