Why $e^0=I$ for exponential operator?

Question

From definition of $e^T=\sum_{k\geq 0} \frac{T^k}{k!}$, so $e^0=0^0$...

So, why $e^0=I$ for exponential operator?

Any help would be appreciated.

Answer 1

The matrix exponential is defined by:

$$\exp(\boldsymbol{A}):= \boldsymbol{I}+\dfrac{1}{1!}\boldsymbol{A}+\dfrac{1}{2!}\boldsymbol{A}^2+\ldots$$

Set $\boldsymbol{A}=\boldsymbol{0}$ to get the result.

A way to understand what $0^0$ means or how we can define a useful value to this expression is to consider the definition: $$0^0:=\lim_{a\to 0}a^a=\lim_{a\to 0}\exp(a\ln a)=\exp(\lim_{a\to 0}a\ln a)=\exp(0)=1.$$

The limit $$\lim_{a\to 0}a\ln a=\lim_{a\to 0}\dfrac{\ln a}{\dfrac{1}{a}}$$

evaluate the last expression by using Bernoulli's limit law (aka L'Hospital's rule).

For a matrix we can also use a similar approach (assuming $\boldsymbol{A}$ is invertible):

$$\boldsymbol{I}=\boldsymbol{A}\boldsymbol{A}^{-1}=\boldsymbol{A}^0.$$

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An even more simple approach is to observe for $b\neq 0$:

$$1=\dfrac{b^a}{b^a}=b^0.$$

It makes sense to define $0^0:=1$ by the previous comment.