Polynomials of linear maps

Hi Professor Kim,

Could you help me with a (potentially trivial) question?

On page 28 of the notes in the proof of Lemma 3.4.80:

fg(T)v = fg(T)(w1 + w2) = gf(T)w1 + fg(T)w2

Why are you able to switch the order of f and g in the final equality when acting on w1? Is it due to w1 being in the kernel of f(T)?

Many thanks
—————————————————————————

Perhaps the notation was a bit misleading. fg(T) refers to the linear map obtained by multiplying the polynomials f(x) and g(x) and then plugging the map T into the place of the variable $x$. So the equality follows from the fact that f(x)g(x)=g(x)f(x).

On the other hand, it is easy to check that this is the same as composing f(T) with g(T) or g(T) with f(T). The reason you can do it in either order is because the only maps appearing are T, and `T commutes with T‘ so to speak. That is, if we had two different maps T and S, we definitely wouldn’t have

f(T)g(S)=g(S)f(T)

in general.

To give a formal proof of the equality

f(T)g(T)=[f(x)g(x)](T),

first do it for f(x)=x^i, which is easy, then just work it out for the general form f(x)=\sum_{i=0}^nc_ix^i of f(x):

[f(x)g(x)](T)=[\sum_{i=0}^nc_ix^ig(x)](T)=\sum_{i=0}^nc_iT^ig(T)

=(\sum_{i=0}^nc_iT^i)g(T)=f(T)g(T).

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