$k[x]$-module and cyclic module over a finite dimensional vector space

Given a finite dimensional vector space $V$ over a field $k$ and a linear transformation $T: V \rightarrow V$ we can make $V$ a $k[x]$-module via the map:

$$(a_{0}+a_{1}x+\cdots+a_{n}x^{n}) \cdot v \mapsto \sum a_{i}T^{i}(v).$$

On page 10 of this file I was reading the following result:

"$V$ is a cyclic $k[x]$-module iff the minimal polynomial and the characteristic polynomial coincide"

Can anyone please show me how to prove this? Or any reference?

Thanks.

• The This in the sentence you quote cannot refer the anything you wrote before it. Maybe if you provide context, we can help. – Mariano Suárez-Álvarez Jun 10 '11 at 21:37
• One direction is easy: every cyclic $K[x]$-submodule of $V$ has dimension at most the degree of the minimal polynomial, so if the minimal polynomial does not equal the characteristic polynomial, no cyclic $K[x]$-submodule of $V$ can equal $V$ (the dimension is too small). – Arturo Magidin Jun 11 '11 at 4:19
• The important point really is the degree of the minimal polynomial, not that it is equal to the characteristic polynomial. – lhf Jun 11 '11 at 15:45

As I mentioned in comments, one direction is easy: if the minimal polynomial is not equal to the characteristic polynomial (up to the leading term), then its degree is stricly smaller than $\dim(V)=n$; then every $T$-cyclic subspace has dimension at most $m$ (the degree of the minimal polynomial), hence cannot equal all of $V$. Thus, if the minimal polynomial does not equal the characteristic polynomial, then $V$ cannot be $T$-cyclic (which means it is not a cyclic $k[x]$-module).

The converse, however, is subtler. Suppose the minimal polynomial equals the characteristic polynomial. That means that for every polynomial $p(t)$ of degree stricly smaller than $n$, $p(T)$ is not identically zero; that is, for every polynomial $p(t)$ of degree strictly smaller than $n$, there exists a vector $v$ such that $p(T)(v)\neq 0$. This does not, however, in and of itself, imply that there is a vector $v$ such that for every polynomial $p(t)$ of degree strictly smaller than $n$ we have $p(T)(v)\neq 0$. Just because for every polynomial there is a vector that works, it does not follow that there is a vector that works for every polynomial.

Instead, we need to work a bit harder. We need to show that $V$ has a $T$-cyclic subspace whose dimension equals that of the minimal polynomial.

Write the minimal polynomial as a product of powers of distinct irreducible polynomials: $$m(t) = \phi_1(t)^{k_1}\cdots\phi_r(t)^{k_r}.$$ The first step is, for each $i$, to find a vector $v_i$ whose annihilator is $\phi_i(t)^{k_i}$ and no smaller power of $t$.

Consider $$K_{\phi_i} = \{v\in V\mid \text{there exists }p\text{ such that }\phi_i(T)^p(v)=0\}.$$ Then $V=K_{\phi_1}\oplus K_{\phi_2}\oplus\cdots\oplus K_{\phi_r}$ (this holds whether or not the minimal polynomial equals the characteristic polynomial).

Now let $S_i$ be the collection of all $p$ such that there exists $v\in K_{\phi_i}$ such that $\phi_i(t)^{p}(v)=0$, but $\phi_i(t)^{p-1}(v)\neq 0$. It is easy to verify that $S_i$ is finite, since we can take an arbitrary basis for $K_{\phi_i}$ and simply take the exponents corresponding to a basis. Let $M$ be the maximum; then $\phi_i(t)^M$ annihilates $K_{\phi_i}$. That means that $$\frac{m(t)\phi_i(t)^M}{\phi_i(t)^{k_i}}$$ annihilates $V$, which means that this polynomial is a multiple of the minimal polynomial. Hence $M\geq k_i$; but it is also easy to check that $\phi_i(t)^{k_i}$ annihilates $K_{\phi_i}$, so $M=k_i$. Therefore, there exists a vector $v_i\in K_{\phi_i}$ such that $\phi_i(t)^{k_i}(v_i)=0$, but $\phi_i(t)^{k_i-1}(v_i)\neq 0$.

Now we have the following:

Theorem. Let $V$ be a vector space, $T$ a linear operator on $V$, and let $w_1$, $w_2$ be two vectors. Suppose that the $T$-annihilators $p_1(t)$ and $p_2(t)$ of $w_1$ and $w_2$ (respectively) are relatively prime. Then the $T$-annihilator of $w_1+w_2$ is $p_1(t)p_2(t)$.

Proof. To show that $p_1(T)p_2(T)$ annihilates $w_1+w_2$ is straightforward, since $p_1(T)p_2(T)=p_2(T)p_1(T)$.

Note that if $W_1$ is the $T$-cyclic subspace generated by $w_1$ and $W_2$ is the $T$-cyclic subspace generated by $w_2$, then $W_1\cap W_2=\{0\}$, since any vector in the intersection must be annihilated by both $p_1(T)$ and $p_2(T)$, so its annihilator must divide their gcd, which is $1$.

Let $f(t)$ and $g(t)$ be polynomials such that $f(t)p_1(t)+g(t)p_2(t)=1$. Then $f(T)p_1(T)$ is the projection of $W_1\oplus W_2$ onto $W_2$: note that $$f(T)p_1(T)(T^i(w_1)) = T^i(f(T)p_1(T)(w_1)) = T^i(f(T)(0)) = 0,$$ and $$f(T)p_1(T)(T^j(w_2)) = T^j\Bigl(\bigl(1 - g(T)p_2(T)\bigr)(w_2)\Bigr)= T^j(w_2).$$

Consider the $T$-cyclic subspace generated by $w_1+w_2$. It is $T$-invariant, so it is also invariant under $f(T)p_1(T)$; hence $w_2 = f(T)p_1(T)(w_1+w_2) = w_2$ lies in the $T$-cyclic subspace generated by $w_1+w_2$, and hence so does $W_1$; since $w_1+w_2$ and $w_1$ lie in the space, so does $w_2$, and therefore it also contains $W_2$. Therefore, we have that the $T$-cyclic subspace generated by $w_1+w_2$ contains $W_1\oplus W_2$, and hence must be equal to it.

Finally, the dimension of $W_1\oplus W_2$ (which is the $T$-cyclic subspace generated by $w_1+w_2$) equals the sum of the degrees of $p_1(t)$ and $p_2(t)$; we already know that $p_1(t)p_2(t)$ annihilates the space, hence is a multiple of the $T$-annihilator of $w_1+w_2$. Since the annihilator has the same degree as $p_1(t)p_2(t)$, and they are both monic, it follows that $p_1(t)p_2(t)$ is the $T$-annihilator of $w_1+w_2$, as desired. QED

Corollary. Let $V$ be a vector space, $T$ an operator on $V$, and let $w_1,\ldots,w_n$ be vectors in $V$ with corresponding $T$-annihilators $p_1(t),\ldots, p_n(t)$. If the $p_i(t)$ are pairwise relatively prime, then the $T$-annihilator of $w_1+\cdots +w_n$ is $p_1(t)\cdots p_n(t)$.

Putting this together, we have:

Theorem. Let $V$ be a finite dimensional vector space, let $T$ be an operator on $V$, and let $m(t)$ be the minimal polynomial of $T$. Then $V$ has a $T$-cyclic subspace of dimension $\deg(m(t))$.

Proof. Write $m(t)= \phi_1(t)^{k_1}\cdots \phi_r(t)^{k_r}$ as a product of powers of pairwise distinct irreducible polynomials. We know that for each $i$ there is a vector $v_i$ whose $T$-annihilator is $\phi_i(t)^{k_i}$. By the theorem above, the $T$-annihilator of $v=v_1+\cdots+v_r$ is $\phi_1(t)^{k_1}\cdots \phi_r(t)^{k_r}=m(t)$, and in particular the $T$-cyclic subspace generated by $v$ has dimension $\deg(m(t))$, as desired. QED

Now it follows that if the minimal polynomial has degree $\dim(V)$, then there exists a vector $v$ such that no polynomial of degree strictly smaller than $m(t)$ will $T$-annihilate $v$, so $v$ is a witness to the fact that $V$ is $T$-cyclic (and hence $k[x]$-cyclic under the defined action).

• thank you for the great answer – user10 Jun 13 '11 at 14:04

There is a clear and detailed proof of this fact in Dummit and Foote's Algebra book, if I'm not mistaken.

Arturo's answer above is complete and self contained. Just thought I would add an answer/comment, in case you know the fundamental theorem of finitely generated modules over a PID(FTFMP):

By FTFMP, $V\cong k[x]/(a_1(x))\times\cdots \times k[x]/(a_m(x))$, where the characteristic polynomial $\chi_T(x)=a_1(x)\times\cdots\times a_m(x)$ and the minimal polynomial $m_T(x)=a_m(x)$.

So if $\chi_T(x)=m_T(x)$, then $V\cong k[x]/(a_m(x))$, a cyclic $k[x]$-module.