I have to show that the derivative of 'the matrix exponential' $exp: \mathbb{C}^{n\times n}\mapsto\mathbb{C}^{n\times n}$ at the zero matrix $0$ is $id_{C^{n\times n}}$, i.e. $exp(0)=id$. The above map isn't given explicitly, so maybe someone can tell me what it does to a matrix $A$? I think the obvious guess would be $$A\mapsto e^A:=\sum_{k=0}^\infty \frac{A^k}{k!}$$, but I wouldn't know how to find the derivation of such a map since there is no variable.

So far I tried considering $$Ax\mapsto e^Ax:=\sum_{k=0}^\infty \frac{(Ax)^k}{k!}$$ instead, which is a map $\mathbb{C}\rightarrow\mathbb{C}$ I suppose.

As derivative I get $Ax\mapsto Ae^{Ax}$, which gives the identity for $x=0$, but $0$ for $A=0$.

Can somebody tell me what I got wrong, and how it is true that $exp(0)=id$?

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    $\begingroup$ $A$ is the variabile. $\endgroup$ – N74 Jun 15 '16 at 18:39

$e^A$ is indeed $\sum_{n=0}^\infty \frac{A^k}{k!}$. To find the derivative at the zero matrix, let $B$ be an arbitrary $n \times n$ matrix which is unit in some norm, let $\epsilon>0$ be a small number, and consider

$$e^{\epsilon B}-e^0=\sum_{k=0}^\infty \frac{\epsilon^k B^k}{k!}-\sum_{k=0}^\infty \frac{0^k}{k!} = \sum_{k=1}^\infty \frac{\epsilon^k B^k}{k!}.$$

(Here by definition $A^0=I$ for any $n \times n$ matrix $A$, including the zero matrix.) Any idea what to do now? (Hint: what is the relative size of the terms?)

  • $\begingroup$ Note: we're using the Frechet derivative here $\endgroup$ – Omnomnomnom Jun 15 '16 at 18:49
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    $\begingroup$ @Omnomnomnom The space is finite dimensional, the distinction between different sorts of derivatives doesn't matter. (Or perhaps I misunderstood the point of your comment?) $\endgroup$ – Ian Jun 15 '16 at 18:51
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    $\begingroup$ @see This definition is not so different from the calculus definition. The reason that there must be some difference is that you can't divide by a vector. So instead you perturb from your starting point going a small distance in some direction and divide by the distance that you went. So here that would be $\frac{e^{\epsilon B}-e^0}{\epsilon}$. Then you send $\epsilon\to 0$. This result depends on the direction $B$ that you went in, but (assuming your function is differentiable in the first place) this dependence is linear. We call the derivative this linear map that $B$ gets plugged into. $\endgroup$ – Ian Jun 15 '16 at 19:46
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    $\begingroup$ @see You probably used essentially the same idea when you looked at gradients in multivariable calculus. The difference is that the derivative of a scalar valued function $f$ of a vector variable $x$ at a point $x_0$ is not really $\nabla f(x_0)$ but rather the map $y \mapsto \nabla f(x_0) \cdot y$, where $\cdot$ is the dot product. Similarly the derivative of a vector valued function of a vector variable is $y \mapsto J(x_0) y$ where $J$ is the Jacobian and the multiplication is ordinary matrix multiplication. $\endgroup$ – Ian Jun 15 '16 at 19:46
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    $\begingroup$ @Ian the point is that this wikipedia page is the one with the nicest explanation of the derivative, as it applies to your answer. $\endgroup$ – Omnomnomnom Jun 15 '16 at 20:31

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