If $\lambda$ is an eigenvalue for a linear operator $T(x)$ and if $g(t)$ is a polynomial, then $g(T(x)) = g(\lambda)x$

Question

Also note that, $x \in V$, where $V$ is a vectorspace over a field $F$ and the co-efficient's of $g(t)$ are in $F$

My attempt:

I proved that $T^m(x) = \lambda^mx$ for any positive integer $m$ in an earlier question.

Using that fact we have, $$g(T(x))= a_m\lambda^mx + .... + a_2\lambda^2x + a_1\lambda^1x + a_0$$

but the constant term doesn't allow $g(T(x)) = g(\lambda)x$ unless $a_0=0$ or $x=1$ as far as I can see.

Am I missing something?

Answer 1

As written this makes no sense: you don't have a multiplication operation on vectors. Presumably what is meant is $g(T)x$ or $(g(T))(x)$, and $x$ is an eigenvector of $T$ with eigenvalue $\lambda$. For that, it is easy to check the desired result. Note that by convention $T^0=I$, so the "constant" term still has an x in it (it just does not have a $T$ in it).