How is the matrix exponential the sum of the identity matrix and the original matrix?

Question

I was watching this YouTube video and at 1:45 the instructor writes the following on the board

$$e^A=A+I$$

How is this true? Is there a proof?

Answer 1

In general, the matrix exponential is defined by $$e^A = \sum_{n = 0}^{\infty}\dfrac{1}{n!}A^n = I + A + \dfrac{1}{2}A^2+\dfrac{1}{6}A^3+\cdots,$$ and so in general $e^A = I+A$ is not true. However, the instructor is showing an example of the matrix exponential for the matrix $$A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}.$$ This particular matrix $A$ satisfies $A^2 = 0$, and thus, $A^n = 0$ for all integers $n \ge 2$. Hence, only the first two terms of the above summation are non-zero, and we get $e^A = I+A$.

Answer 2

This is not true in general. In the video the example is

$$A=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$ for which $A^n = 0 $ for $n \ge 2$. Hence the conclusion $e^A = \sum_{k=0}^\infty \frac{A^n}{n!}=I+A$.