Is every eigenvector of $g(T)$ is also Eigenvector of $T$?

Question

Let's say i have a linear transformation $T: V \longrightarrow V$ and a polynomial $g$.

I know that every eigenvector of $T$ with eigenvalue $\alpha$ is an eigenvector of $g(T)$ with eigenvalue $g(\alpha)$. Does it necceserily mean that every eigenvector of $g(T)$ is also an eigenvector of $T$?

Answer 1

Suppose that $T(x,y)=(x,-y)$ and that $g(x)=x^2$. Then $g(T)(x,y)=(x,y)$ and then every non-null vector of $k^2$ (where $k$ is the field that you are working with) is an eigenvector of $g(T)$. But it is not true that every non-null vector of $k^2$ is an eigenvector of $T$.

Answer 2

Not true if $g$ is a constant. Also not true for many other polynomials including $g(x)=x^{2}$.

Answer 3

Consider what happens if $g$ is the characteristic polynomial of $T$. Then every (non-zero) vector is an eigenvector of $g(T)$ (with eigenvalue $0$), by the Cayley-Hamilton theorem. And unless $T$ is a scalar multiplication, $T$ has non-eigenvectors.