# Linear Operators, Eigenvalues

1. Oct 29, 2006

### frederick

If A & B are linear operators, and AY=aY & BY=bY, what is the relationship between A & B such that e^A*e^B=e^(A+B)?? --where e^x=1+x+x^2/2+x^3/3!+...+x^n/n!

2. Oct 30, 2006

### dicerandom

Hint: write it out. And remember to watch out for which side you're multiplying on with A or B since, in general, $AB \neq BA$.

3. Nov 1, 2006

### dextercioby

Since Y is a common eigenvector for A and B (assuming of course they don't have a continuous spectrum), then A and B commute. Then using Baker-Campbell-Hausdorff formula one sees that "A and B commute" is enough to have $e^{A}e^{B}=e^{A+B}$.

Daniel.

4. Nov 1, 2006

### Hurkyl

Staff Emeritus
What am I missing? Consider the matrices:

$$\left(\begin{array}{cc} 3 & 0 \\ 1 & 1 \\ \end{array}\right)$$ and $$\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \\ \end{array}\right)$$

These share a common eigenvector $[0 1]^T$, but don't commute.

5. Nov 1, 2006

### dextercioby

Yes, sloppy me, the theorem goes: 2 linear operators commute iff they share a COMPLETE system of common eigen vectors.

Daniel.