Fréchet derivative of the (matrix) exponential function

$$
D\exp(x)u = \int_0^1 e^{sx}ue^{(1-s)x}\,ds.
$$ This intriguing formula expresses the derivative of the exponential map on a Banach algebra as an integral. In particular, using “matrix calculus” notation we have the formula $$
d\exp(X)= \int_0^1 e^{sX}(dX)e^{(1-s)X}\,ds
$$ when \(X\) is a square matrix. As we’ll see, this is not too hard to prove.

We will assume that all Banach algebras are unital.

Definition. Let \(E\) be a Banach algebra. If \(x\in E\), the exponential of \(x\) is $$
\exp(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!},
$$ which converges absolutely for all \(x\). Thus we have a map \(\exp:E\to E\), called the exponential function.

The usual rules for power series apply. In particular, we can differentiate term by term inside the radius of convergence, which is infinite for the exponential function. Before doing this, we need a lemma (the proof is at the end of the post).

Lemma 1 (Power rule). Let \(E\) be a Banach algebra, let \(n\ge 0\), and let \(p_n:E\to E\) be the map defined by \(p_n(x)=x^n\). Then \(Dp_n(x)\) is the linear map given by $$
Dp_n(x)u=\sum_{k=0}^{n-1} x^kux^{n-k-1}.
$$ In particular, if \(E\) is commutative then \(Dp_n(x)\) is given by $$
Dp_n(x)u=nx^{n-1}u.
$$

Applying this lemma to the power series for \(\exp\) gives $$
D\exp(x)u=\sum_{n=1}^\infty \frac{1}{n!} \sum_{k=0}^{n-1}x^kux^{n-k-1}.\tag{*}
$$ Notice that when \(u\) commutes with \(x\), we have \(D\exp(x)u=\exp(x)u=u\exp(x)\). We also need another lemma (again, the proof is at the end of the post):

Lemma 2. For \(m,n\ge 0\), we have $$\int_0^1 s^m(1-s)^n\,ds=\frac{m!n!}{(m+n+1)!}.$$

Evaluation

Now we can evaluate the integral given at the beginning of the post. We have \begin{align}
\int_0^1 e^{sx}ue^{(1-s)x}\,ds &= \int_0^1 \sum_{m=0}^\infty\frac{s^m x^m}{m!}u\sum_{n=0}^\infty\frac{(1-s)^n x^n}{n!}\,ds \\
&= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{x^m u x^n}{m!n!}\int_0^1 s^m(1-s)^n\,ds \\
&= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{x^m u x^n}{(m+n+1)!},
\end{align} which is clearly equal to (*). (The rearrangements are valid because the infinite series are all absolutely convergent.) This proves our formula!

Proofs

Proof of Lemma 1. We use induction on \(n\). The case \(n=0\) is clear, so suppose that the result holds for \(n-1\). Since \(p_n(x)=xp_{n-1}(x)\), the product rule shows that \(Dp_n(x)\) maps \(u\) to \begin{align}
up_{n-1}(x)+xDp_{n-1}(x)u &= ux^{n-1}+x\sum_{k=0}^{n-2}x^kux^{n-k-2} \\
&= ux^{n-1}+\sum_{k=1}^{n-1}x^kux^{n-k-1} \\
&= \sum_{k=0}^{n-1} x^kux^{n-k-1}.
\end{align} \(\square\)

Proof of Lemma 2. We use induction on \(n\). The case \(n=0\) is obvious. Suppose the formula holds for \(n-1\). We have \begin{align}
\int_0^1 s^m(1-s)^n\,ds &= \int_0^1 s^m(1-s)^{n-1}(1-s)\,ds \\
&= \int_0^1 s^m(1-s)^{n-1}\,ds- \int_0^1 s^{m+1}(1-s)^{n-1}\,ds \\
&= \frac{m!(n-1)!}{(m+n)!} – \frac{(m+1)!(n-1)!}{(m+n+1)!} \\
&= \frac{m!(n-1)!(m+n+1)-(m+1)!(n-1)!}{(m+n+1)!} \\
&= \frac{m!n!}{(m+n+1)!}.
\end{align} \(\square\)

Leave a Reply