What are the conditions for commuting linear maps?

[Marin]

Hi all!

I was wondering what conditions must two linear maps obey in order to commute?

If they are described by two matrices A and B, then one condition would be:

AB-BA=0

but what if we don't know the matrices, so we cannot compute AB adn BA? How is one supposed to proceed, is there a more general condition?



And another question: Suppose A and B commute. Will they still commute if I change the basis, I mean is the commutation coordinate independent?


thanks a lot for the help,

marin

 

[ThirstyDog]

I hope this answers in part your questions:

If you have two linear maps
f: V \rightarrow W \mbox{ and } g: V' \rightarrow W'
if you want the two maps to commute then what you are saying is you want the composition of maps to satisfy
f \circ g = g \circ f \hspace{0.3cm} \Rightarrow f(g(v)) = g(f(v)).
If this is the case then you need that
V = W = V' = W'
for the composition to make sense. In terms of matrices this implies that they are square.

In response to your question about changing the basis as
V = W = V' = W'
you must perform the basis transformation on each space and if this happens then commuting matrices will still commute under a change of basis.

 

[Marin]

ok, so I need my linear maps to be endomorphisms and the matrices to be square :)

but this is not enough, since there are plenty of square matrices which do not commute.

So what else do I need?

 

[ThirstyDog]

The only thing you will need is
f \circ g = g \circ f \hspace{0.3cm} \Rightarrow f(g(v)) = g(f(v)).

If the linear maps are matrices, A and B, then \circ is just the usual matrix multiplication which gives
AB = BA.
This is exactly what you had. I don't think that there is anything else needed and I can't think of any simpler equivalent conditions.

 

[eok20]

One condition is that two linear maps commute if and only if they commute on all vectors of some basis. To see that commutation is independent of basis, note that in a new basis, the transformation A becomes PAP^{-1} and B becomes PBP^{-1} for some change of basis matrix P. If A and B commute then (PAP^{-1})(PBP^{-1}) = PABP^{-1} = PBAP^{-1} = (PBP^{-1})(PAP^{-1}).