Are Linear Operators Commutative When They Share Common Eigen Vectors?

[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!

 

[dicerandom]

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

 

[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.

 

[Hurkyl]

then A and B commute.

What am I missing? Consider the matrices:

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

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

 

[dextercioby]

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

Daniel.