On exponentials of matrices

Question

How can I prove that $|e^{A}|\leq e^{|A|}$ where $|\cdot|$ means the operator norm of matrices?

I guess i'm having trouble with the definition!

A is a square (complex or real) matrix.

Answer 1

It's wrong!

Let $A=\begin{pmatrix}100&0\\0&0\end{pmatrix}$. Then $|A|=0$ so $e^{|A|}=1$. But $$ e^{A}=\begin{pmatrix}e^{100}&0\\0&1\end{pmatrix} $$ so $|e^{A}|=e^{100}>1=e^{|A|}$.

Answer 2

Let $\|.\|$ be a sub multiplicative norm on $\Bbb C^{n\times n}$ (ie. $\|AB\|\le\|A\|\|B\|$). The matrix exponential for an element $A\in\Bbb C^{n\times n}$ is defined as follows. $e^A:=\sum_{k=0}^\infty\frac{A^k}{k!}$. To show the inequality, we first show $\|A^k\|\le\|A\|^k$ by induction. Then:

$\|e^A\|=\|\sum_{k=0}^\infty\frac{A^k}{k!}\|=\|\lim_{n\to\infty}\sum_{k=0}^n\frac{A^k}{k!}\|=\lim_{n\to\infty}\|\sum_{k=0}^n\frac{A^k}{k!}\| \le\lim_{n\to\infty}\sum_{k=0}^n\|\frac{A^k}{k!}\| \le\lim_{n\to\infty}\sum_{k=0}^n\frac{\|A\|^k}{k!}=e^{\|A\|}$