3
$\begingroup$

Let $$ T: V \rightarrow V$$ be a linear operator on a vector space $V$ over a field $F$ and with a basis $$ B = (\mathbf{u}_{1}, \cdots, \mathbf{u}_{n}).$$ Let $$ f(x), g(x) \in F[x]$$ be polynomials. Then

$$ f(T)g(T) = g(T)f(T).$$

Does that mean if i am given a linear operator $T:V \to V$, and another map $T - \lambda I_V$, i can say both maps commute with each other because of the above theorem.

I say : $$f(x) = x,~~~ g(x) = x - \lambda$$ Then $$f(T) = T,~~~ g(T) = T - \lambda I_V$$

and hence they commute. Moreover, if $g(x) = (x-\lambda)^n$, then $h(T) = (T-\lambda I_V)^n$ and subsequently by the above theorem, we have $$T \circ (T- \lambda I_V)^n = (T - \lambda I_V)^n \circ T$$

Can anyone clarify this? I am working on the proof of Jordan normal form.

$\endgroup$
3
  • 2
    $\begingroup$ Yes, $T$ commutes with powers of $T-\lambda I$. $\endgroup$ Commented Nov 25, 2017 at 6:13
  • 1
    $\begingroup$ Yes, but is my line of logic correct in showing that they commute ? $\endgroup$
    – nan
    Commented Nov 25, 2017 at 6:14
  • 2
    $\begingroup$ Have a little confidence! $\endgroup$ Commented Nov 25, 2017 at 6:17

1 Answer 1

4
$\begingroup$

The linear operators commute if they are the polynomials of the same operator. So yes .What you said is true.

$\endgroup$
1
  • 1
    $\begingroup$ How do we prove that linear operators commute if they are the polynomials of the same operator? $\endgroup$ Commented Oct 9, 2023 at 14:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .