## 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
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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 $f(x)$:

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

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