Show that $e^{A}$ makes sense, $A$ be an $n \times n$ real matrix

Question

I have to solve:

Let $A$ be an $n \times n$ real matrix. The exponential of $A$ is defined as $$ e^{A}:=I+A+\frac{A^{2}}{2 !}+\frac{A^{3}}{3 !}+\cdots $$ provided each entry of the matrix converges.

1. Show that $e^{A}$ makes sense
2. and $P$ is an invertible matrix show that $P e^{A} P^{-1}$ is also an exponential matrix.

I have solved Part (2) as :
This follows by noting that $$ \begin{aligned} \left(T B T^{-1}\right)^{k} &=\left(T B T^{-1}\right)\left(T B T^{-1}\right) \cdots\left(T B T^{-1}\right) \\ &=T B\left(T^{-1} T\right) B\left(T^{-1} T\right) \cdots\left(T^{-1} T\right) B T^{-1} \\ &=T B^{k} T^{-1} \end{aligned} $$

then yields $$ \begin{aligned} e^{P^{-1} A P} &=I+P^{-1} A P+\frac{\left(P^{-1} A P\right)^{2}}{2 !}+\cdots \\ &=I+P^{-1} A P+P^{-1} \frac{A^{2}}{2 !} P+\cdots \\ &=P^{-1}\left(I+A+\frac{A^{2}}{2 !}+\cdots\right) P=P^{-1} e^{A} P \end{aligned} $$

• I don't know how to solve part (1) i.e Show that $e^{A}$ makes sense ?

Answer 1

Part 1 means to ask that you show the series actually converges for any A. (I always see this done after the exponential is introduced, using absolute convergence.)

Answer 2

As Keshav notes, you can prove that the definition of the exponential makes sense as long as you can show that the infinite series converges for any $A$. The obvious way to do this is to compare it to the exponential series for real numbers, which you presumably already know converges.

For a matrix $A$, denote by $|A|$ the operator norm of $A$. This has the property that $|A^k| \le |A|^k$. Therefore, if we take a finite sum:

$$\sum_{i=0}^N \left|A^i/i!\right| \le \sum_{i=0}^N|A|^i/i! \to e^{|A|}$$ so that the sum converges absolutely.