# Derivative of matrix exponential of linear combination

Let's say $$t$$ is a real parameter and $$\textbf{A}$$ is $$n \times n$$ matrix. I know that

$$\frac{d}{dt} \exp(t\textbf{A}) = \textbf{A} \exp(t\textbf{A})$$

But what if there are multiple parameters $$t_1, ..., t_n$$ and multiple matrices $$\textbf{A}_1, \dots , \textbf{A}_n$$ that don't commute, does the same relation, namely

$$\frac{\partial}{\partial t_k}\exp \left(\sum_{i=1}^n t_i \textbf{A}_i \right) = \textbf{A}_k \exp \left(\sum_{i=1}^n t_i \textbf{A}_i\right)$$

still hold? If not, is there a closed form for the derivative?

• $\exp(tA+sB)-\exp(tA)\neq\exp(tA)(\exp(sB)-I)$ if $A$ and $B$ do not commute, so no, it doesn't. $\textbf{A}_k \exp(\sum_{i=1}^n t_i \textbf{A}_i)\neq \exp(\sum_{i=1}^n t_i \textbf{A}_i)\textbf{A}_k$ either, as it was with a single generator. – Conifold Jan 10 at 10:09
• This can also be simply seen from the BCH formula – Mathphys meister Jan 10 at 11:00
• Ok, but is there a closed-form result for the derivative in this case? – xaxa Jan 10 at 11:28
• There is no closed form unless you have some simple commutation relations among the generators. – Conifold Jan 10 at 23:42

## 3 Answers

PARTIAL ANSWER.

If there is a general formula for $$\partial_t e^{tX+sY}, \qquad \partial_s e^{tX+sY},$$ then it surely descends from the Baker-Campbell-Hausdorff formula, which is known to be rather complicated. So, I will only consider a simpler special case.

In the special case of $$X$$ and $$Y$$ commuting with the commutators, that is $$\tag{1}[X, [X, Y]]=[Y, [X, Y]]=0,$$ the Baker-Campbell-Hausdorff formula simplifies to $$\tag{2} e^{tX+sY+\frac{ts}{2}[X, Y]}=e^{tX}e^{sY},$$ and since, by (1), $$tX+sY$$ commutes with $$[X, Y]$$,
$$e^{tX+sY}=e^{tX}e^{sY}e^{-\frac{ts}{2}[X, Y]},$$ from which we infer $$\begin{split} \partial_t e^{tX+sY} &= Xe^{tX}e^{sY}e^{-\frac{ts}{2}[X,Y]} - e^{tX}e^{sY}\left(\frac{s}{2}[X, Y]\right)e^{-\frac{ts}{2}[X,Y]}\\ &=(X-\frac{s}{2}[X, Y])e^{tX+sY}, \end{split}$$ where we used (1), which implies that $$[X,Y]$$ commutes with $$e^{tX}$$ and with $$e^{sY}$$.

Reversing the roles of $$X$$ and $$Y$$, and using that $$[Y, X]=-[X, Y]$$, we have $$\partial_s e^{tX+sY} = (Y+\frac{t}{2}[X, Y])e^{tX+sY}.$$

In particular, we see that in general it is not true that $$\tag{!!} \partial_t e^{tX+sY} = X e^{tX+sY}.$$

SOME WORDS ON THE GENERAL CASE.

If we drop the assumption (1), then the Baker-Campbell-Hausdorff formula becomes much more complicated; $$\tag{3} e^{tX+sY+\frac{ts}{2}[X, Y]+\frac1{12}\left( t^2s[X,[X,Y]] - ts^2[Y, [X,Y]]\right)+\ldots} =e^{tX}e^{sY}.$$ I don't know how to pass from (3) to a formula of the form $$e^{tX+sY}=e^{tX}e^{sY}e^{\Phi(t, s)},$$ which is what we need for the above computation.

• Thanks! In this case, one might consider function $\exp(tA+B)$ with B and A not commuting. Then we have $\exp((t+dt)A+B) = \exp(tA+B + dt\,A)$, so we cam use BCH formula with $X=tA+B$ and $Y=dt\,A$, $[X,Y]=[B,A]dt$. This might be helpful becaise $Y$ is linear in $dt$ and we don't need higher orders of $dt$ when calculating derivatives, so we might only consider commutators like $[X,Y]$, $[X,[X,Y]]$ and so on. But still... I don't see where to go with that. – xaxa Jan 10 at 14:22
• If you do that kind of manipulations you will end up with some formula. But I am not sure that it will be simple enough to be usable. – Giuseppe Negro Jan 10 at 14:30
• You can get an expression of the type written at the end from the Zassenhaus formula, see also multivariable Zassenhaus formula. – Conifold Jan 10 at 23:57
• @Conifold: Thank you, that's exactly what is needed here. – Giuseppe Negro Jan 13 at 10:13

As for analytical functions in one dimension, one can define a matrix function for square matrices $$\mathbf{X}$$ by an infinite series: $$f(\mathbf{X}) = \sum_{n=0}^\infty c_n \mathbf{X}^n$$ assuming the limit exists and is finite. If the coefficients $$c_n$$ fulfils $$\sum c_n x^n <\infty$$, then one can prove that the above series exists and is finite

source: The matrix cookbook

So we know that $$\exp(t\mathbf{A})$$:

$$\exp(t\mathbf{A}) = \sum_{n=0}^\infty \frac{t^n \mathbf{A}^n}{n!}$$

and thus

$$\frac{\textrm{d}}{\textrm{d}t}\exp(t\mathbf{A}) = \sum_{n=1}^\infty \frac{n t^{n-1} \mathbf{A}^n}{n!} = \sum_{n=0}^\infty \frac{t^{n} \mathbf{A}^{n+1}}{n!}=\textbf{A}\,\exp(t\mathbf{A}) = \exp(t\mathbf{A})\,\textbf{A}$$

The OP is interested in

$$\frac{\partial}{\partial t_p}\exp\left\{\sum_{i=1}^nt_i\mathbf{A}_i\right\}.$$

This is not as straightforward as it seems. To explain this you need to understand that:

$$\exp(\mathbf{A} + \mathbf{B}) = \exp(t\mathbf{A})\,\exp(t\mathbf{B}),\quad\textrm{if}\quad \mathbf{AB}=\mathbf{BA}$$

If $$\mathbf{A}$$ and $$\mathbf{B}$$ do not commute, this is not the case. You can see this from the simple expansion:

$$\exp(\mathbf{A} + \mathbf{B}) = \sum_{n=0}^\infty \frac{t^n (\mathbf{A}+\mathbf{B})^n}{n!}$$

Only if $$\mathbf{A}$$ and $$\mathbf{B}$$ commute, you can use the binomial expansion, in the other case it becomes really messy. To give an example, the second term becomes:

\begin{align} \exp(\mathbf{A} + \mathbf{B}) =& \mathbf{I} + \mathbf{A}+\mathbf{B} \\ & + \frac{1}{2}\left(\mathbf{A}^2 + \mathbf{AB} + \mathbf{BA} + \mathbf{B}^2\right) \\ & + \frac{1}{3!}\left(\mathbf{A}^3 + \mathbf{A}^2\mathbf{B} + \mathbf{A}\mathbf{B}\mathbf{A} +\mathbf{B}\mathbf{A}^2 + \mathbf{B}^2\mathbf{A} + \mathbf{B}\mathbf{A}\mathbf{B} +\mathbf{A}\mathbf{B}^2+ \mathbf{B}^3\right)\\ &+\ldots \end{align}

So, in the OP, if $$\mathbf{A}_p\mathbf{A}_q = \mathbf{A}_q\mathbf{A}_p$$ for all combinations of $$p$$ and $$q$$, then

$$\frac{\partial}{\partial t_p}\exp\left\{\sum_{i=1}^nt_i\mathbf{A}_i\right\} = \textbf{A}_p\,\exp\left\{\sum_{i=1}^nt_i\mathbf{A}_i\right\} = \exp\left\{\sum_{i=1}^nt_i\mathbf{A}_i\right\}\,\textbf{A}_p$$

If the matrices do not commute, i.e. $$\mathbf{A}_p\mathbf{A}_q \ne \mathbf{A}_q\mathbf{A}_p$$ for any combination of $$p$$ and $$q$$, it quickly blows up.

• Thanks. I assume that my question is really equivalent to $d/dt \exp(tA+B)$ where $B$ doesn't depend on $t$ and $[A,B] \ne 0$. Given that $B$ doesn't depend on t, is it possible to simplify the series for derivative? I don't see how, but maybe I'm missing something or maybe there is a known theorem for such a case. – xaxa Jan 10 at 11:33
• Do you know something about higher order commutators. In this case the Baker-Campbell-Hausdorff formula becomes very useful. – kvantour Jan 10 at 11:38

Let \eqalign{ t&=t_i,\quad A=A_i,\quad B=\sum_{k\ne i} t_kA_k \\ G(t) &= \frac{d}{dt}\exp(B+tA) \\ } To evaluate $$G(t)$$ at $$t=0$$, use the block-triangular method. \eqalign{ \exp\left(\left[\matrix{B&A\\0&B}\right]\right) &= \left[\matrix{\exp(B)&G(0)\\0&\exp(B)}\right] \\ } To evaluate $$G$$ at $$t=s,\,$$ shift the definition of $$B\to (B+sA)$$

Define block-analogs of the standard basis vectors $$e_1=\pmatrix{1\\0},\quad e_2=\pmatrix{0\\1}$$ by replacing the {$${0,1}$$} elements with the $$n\times n\,$$ {zero, identity} matrices $$E_1=\pmatrix{I\\0},\quad E_2=\pmatrix{0\\I} \in{\mathbb R}^{2n\times n}$$ Then would this qualify as a closed-form solution? \eqalign{ G(s) &= E_1^T\,\exp\left(\left[\matrix{B+sA&A\\0&B+sA}\right]\right)\,E_2 \\ }

• Oh this is so cool! And the paper you cite is amazing, thank you. I need some time to meditate around this and see if I can get anything useful in my case – xaxa Jan 14 at 18:44