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The question is as the Title stated:

I picked a very easy example. However, I am afraid, I am missing something.

The two matrices that I picked are $$A = \begin{bmatrix}1& 0 \\ 0 & 0\end{bmatrix}\quad \text{and}\quad B =\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$

My reasoning is that $$ \exp(A) = \begin{bmatrix}e^1& 0 \\ 0 & 0\end{bmatrix}\quad\text{and}\quad \exp(B) =\begin{bmatrix}0 & 0 \\ 0 & e^1\end{bmatrix}$$

$$\exp(A)\exp(B) = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$ and $$exp(A+B) = \begin{bmatrix} e^1 & 0 \\ 0 & e^1\end{bmatrix}.$$

Please check it for me

Yiorgos S. Smyrlis
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afsdf dfsaf
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    You are not computing matrix exponential correctly: https://en.wikipedia.org/wiki/Matrix_exponential – vadim123 Jan 26 '14 at 16:40
  • I am sorry...the question should be finding an example to show that they are not equal – afsdf dfsaf Jan 26 '14 at 17:06
  • I just changed my example =, please take a look again. – afsdf dfsaf Jan 26 '14 at 17:10
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    $exp(A) = \begin{pmatrix}e&0\0&1\end{pmatrix}$, $exp(B) = \begin{pmatrix}1&0\0&e\end{pmatrix}$, so $exp(A+B) = \begin{pmatrix}e&0\0&e\end{pmatrix}=exp(A)exp(B)$. Try taking $A$ and $B$ that don't commute. – TZakrevskiy Jan 26 '14 at 17:14
  • @afsdfdfsaf, the common theme to all the answers is that you don't take matrix exponentials by simply exponentiating each entry. Your first step is to learn how; start with my link. Only then are you ready to answer questions about matrix exponentiation. – vadim123 Jan 26 '14 at 17:19
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    Please do not change the question to the opposite after having received an answer. The answer is now useless. – Dietrich Burde Jan 26 '14 at 18:18

1 Answers1

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In your example $$A = \begin{bmatrix}0& 0 \\ 1 & 0\end{bmatrix}\quad \text{and}\quad B =\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$ we have that $A^2=A^k=0$, for all $k\ge 2$, and $B^k=B$, for all $k\ge 1$, and therefore $$ \exp A=I +A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!}+\cdots=I+A= \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix}, $$ while $$ \exp B=I +B+\frac{B^2}{2!}+\cdots+\frac{B^n}{n!}+\cdots=I+B+\frac{B}{2!}+\cdots+\frac{B}{n!}+ \cdots=I+(\mathrm{e}-1)B =\begin{bmatrix}1 & 0 \\ 0 & \mathrm{e}\end{bmatrix}.$$ and $$\exp(A)\exp(B) = \begin{bmatrix}1 & 0 \\ 1 & \mathrm{e}\end{bmatrix}\quad\text{and}\quad \exp(A+B) = \begin{bmatrix}1 & 0 \\ \mathrm{e}-1 & \mathrm{e}\end{bmatrix}.$$

Thus your example WORKS!

Note. A characterization of the commutativity of the exponential of matrices is provided by:

Theorem. Let $A,B\in\mathbb R^{n\times n}$. Then $$ \mathrm{e}^{tA}\mathrm{e}^{tB}=\mathrm{e}^{t(A+B)}, $$ for all $t\in\mathbb R$, if and only if $AB=BA$.

For a proof Theorem 5 in these notes.

Yiorgos S. Smyrlis
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