If a is an eigenvalue of $f(T)$, then $a=f(c)$ for some eigenvalue c of T

Question

I have seen many proofs of this property given that the field is algebraically closed, because then you can simply do: $f(x)-a=a_n(x-r_1)(x-r_2)....(x-r_n)$. Since for some $\alpha$, $f(T)(\alpha)=a\alpha$,$(f(x)-a)(\alpha)=0$, so $(a_n(T-r_1)(T-r_2)....(T-r_n))(\alpha)=0$, this means $(T-r_i)(y)=0$ for some $i,y$. so $r_i$ is an eigenvalue of T. We also know $r_i$ is a root of $f(x)-a$. so $f(r_i)=a$ for some eigenvalue $r_i$.

I am wondering if instead of the condition that the field is algebraically closed, we only know that T is triangularizable, can the statement still be proven? I know if T is triangularizable, both the characteristic polynomial and minimal polynomials split. However, this does not say anything about f. Any ideas?

Answer 1

Let $T$ be a triangularizable linear operator on a finite dimensional vector space $V$ over $F.$ Let $f$ be a polynomial over $F.$ Then $a \in F$ is an eigenvalue of $f(T)$ iff $a=f(c)$, where $c$ is a eigenvalue of $T$.

One direction is easy. Here is a proof of the other direction.

Take a basis $B$ for $V$ such that $T$ is represented by a triangular matrix with respect to $B$.

Then all powers of $T$ are represented by a triangular matrix with respect to $B$.

Thus, $f(T)$ is represented by a triangular matrix with respect to $B$.

The diagonal entries in the matrix of $f(T)$ are of the form $f(c)$, where $c$ is the corresponding diagonal entry in the matrix of $T$.

The diagonal entries in a triangular matrix are exactly its eigenvalues. This is the key point.

Therefore, the eigenvalues of $f(T)$ are exactly of the form $f(c)$, where $c$ is an eigenvalue of $T$.