# First-order ODEs, matrix exponentials, and det(exp)

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. $$\square$$

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