I need to prove the following Lemma, which is then used in the proof of the Cyclic Decomposition theorem in "Linear Algebra", by K. Hoffman and R. Kunze.
Let $T$ be a linear operator on a finite-dimensional vector space $V$ over the field $\mathbb{F}$. If $\beta$ and $\mu$ have the same $T$-annihilator, then for any polynomial $f$, $f(T)\beta$ and $f(T)\mu$ have the same $T$-annihilator.
For reference, the $T$-annihilator of a vector $\beta$ is defined as the monic generator of the ideal of all polynomials $f$ such that $f(T)\beta=0$.
Thank you for your time
Best Answer
Update: So I gave it some more thought and came up with the following:
Let $p_{\alpha }$ be the T-annihilator of ${\alpha }$ and $p_{\gamma }$ the T-annihilator of ${\gamma }$
Let h be a polynomial that annihilates f(T)${\alpha }$, then:
$h(T)(f(T){\alpha }$)=0 => $(hf)(T){\alpha }$=0 => $p_{\alpha }$ divides hf, i.e. there exists a polynomial q with $qp_{\alpha }=hf$, and since $p_{\alpha }$=$p_{\gamma }$ it follows that $hf=qp_{\gamma }$
Therefore, $hf(T){\gamma }=(qp_{\gamma })(T){\gamma }=q(T)(p_{\gamma }(T){\gamma })=0$ by definition of $p_{\gamma }$
Then h annihilates $f(T){\gamma }$
By the same reasoning, if a polynomial $h$ annihilates $f(T){\gamma }$, it annihilates $f(T){\alpha }$.
Then the ideal of polynomials which annihilate $f(T){\gamma }$ is equal to the ideal of the polynomials which annihilate $f(T){\alpha }$, and by the uniqueness of the monic generator of the ideal we have than the T-annihilator of $f(T){\alpha }$ equals the T-annihilator of $f(T){\gamma }$