Proving e^Ae^B=e^{A+B} for Commuting Matrices

[kreil]

Homework Statement

Show, by series expansion, that if A and B are two matrices which do not commute, then $e^{A+B} \ne e^Ae^B$, but if they do commute then the relation holds.

Homework Equations

$$e^A=1+A$$
$$e^B=1+B$$
$$e^{A+B}=1+(A+B)$$

The Attempt at a Solution

Assuming that the first 2 terms in the expansion is sufficient to use here (is it?), I got the following:

$$e^Ae^B=(1+A)(1+B)=1+A+B+AB$$

This would be equal if the AB term were not tacked on the end...does this term somehow become zero when the two matrices commute? If I'm on the wrong track please let me know.

Josh

 

[kreil]

I think I figured it out after a little thought...

$$e^Ae^B=(1+A)(1+B)=1+A+B+AB-BA=1+A+B+[A,B]$$

So if A and B commute, [A,B]=0 and the relation holds, else [A,B] does not equal zero and $e^{A+B} \ne e^Ae^B$

 

[baba89]

ua,

The statement you are trying to prove is a well-known property of commuting matrices. It states that if A and B are two matrices that commute, meaning AB=BA, then the exponential of their sum is equal to the product of their exponentials. In other words, e^Ae^B=e^{A+B}. This can be easily shown using the definition of the exponential function for matrices, which is usually given as a power series:

e^X=I+X+X^2/2!+X^3/3!+...

Where I is the identity matrix. Using this definition, we can expand the left-hand side of the equation e^Ae^B and get:

e^Ae^B=(I+A+A^2/2!+...)(I+B+B^2/2!+...)

Multiplying this out, we get:

e^Ae^B=I+(A+B)+(A^2+B^2+A*B)/2!+...

We can see that the terms involving A and B are grouped together and all the terms involving their products are divided by the same factorial. This is because A and B commute, so their products will also commute and can be combined. This is not the case when A and B do not commute, as you can see by expanding e^{A+B} using the same method.

So to answer your question, the AB term does not become zero when A and B commute, but it can be combined with other terms to give the desired result. This property of commuting matrices is important in many areas of mathematics, including linear algebra and differential equations. It allows us to simplify calculations and make certain problems easier to solve.