Degree of minimum polynomial at most $n$ without Cayley-Hamilton?

Question

Let $T$ be a linear transformation of an $n$-dimensional vector space $V$ over a field $k$. It's pretty easy to define the minimum polynomial of $T$ and make sure its degree is between $1$ and $n^2$, inclusive.

Observe $ I = \{ p(x) \in k[x] : p(T) =0\}$ is an ideal in $k[x]$. Indeed, $I$ is the kernel of the evaluation homomorphism $\mathrm{eval}_T: k[x] \to \mathrm{End}(V)$. Notice also that:

• $\mathrm{eval}_T$ is unital homomorphism, so $I$ is a proper ideal.
• The $n^2 + 1$ transformations $I, T,T^2,T^3,\ldots, T^{n^2}$ must be linearly dependent, since $\mathrm{dim}(\mathrm{End}(V)) = n^2$, so there exist scalars $a_0,\ldots,a_{n^2}$, not all zero, such that $a_0I + a_1 T + \ldots + a_{n^2}T^{n^2} = 0$, whence the nonzero polynomial $p(x) = a_0 + a_1 x + \ldots + a_{n^2}x^{n^2}$ belongs to $I$.

Since $k[x]$ is a p.i.d., we may define the minimum polynomial $m(x)$ of $T$ to be the monic generator of the ideal $I$. By the preceding two observations, we have $1 \leq \mathrm{deg}(m(x)) \leq n^2$.

Now, of course, we know that the degree of $m(x)$ actually satisfies $1 \leq \mathrm{deg}(m(x)) \leq n$. One way to see this is to use the Cayley-Hamilton theorem which shows that the characteristic polynomial $c(x)=\det(xI - T)$, whose degree is $n$, annihilates $T$, whence $m(x)$ divides $c(x)$.

Question: Is there another way to see that $T$ is annihilated by a polynomial of degree $\leq n$ which does not require use of the characteristic polynomial?

Answer 1

Proposition 1.10 of these middle-brow linear algebra notes reproduces an inductive proof of M.D. Burrow.

Here "middle-brow" means: no modules, no tensor products, no change of ground field or use of an algebraically closed ground field until the aftermath of all the main results*; but it does use arithmetic in the univariate polynomial ring $F[t]$ and also quotient spaces.

*: For the majority of the notes, that a matrix has any eigenvalues at all is a pleasant special case!

Answer 2

I like this approach for really explaining why the result is true. For each vector $v$, define the minimal polynomial of $T$ on $v$ to be the monic polynomial of least degree in $T$ that annihilates $v$.

Lemma: If $P$ is the minimal polynomial of $T$ on the vector space $V$, there is some vector $v$ such that the minimal polynomial of $T$ on $v$ equals $P$.

proof sketch: decompose $V$ as a direct sum of kernels of powers of irreducible factors of the minimal polynomial. The lemma follows easily for each summand; then add to get a vector $v$ for all of $V$.

Then since evaluation at $v$ is a linear map from the polynomial ring to the vector space, it follows that the kernel has codimension $\le \dim(V)$, hence the minimal polynomial of $T$ on $v$ has degree $\le \dim(V)$.

Answer 3

Take the tensor product of the vector space with the field of rational functions $k(x_1,x_2,\ldots,x_n)$ and consider the vector $v=(x_1,x_2,\ldots,x_n)$. The vectors $v,Tv,\ldots,T^nv$ are linearly dependent over the field of rational functions, and by clearing denominators we see that there are polynomials $p_0,p_1,\ldots,p_n$, at least one of which is nonzero, such that $$p_0v+p_1Tv+\cdots+p_nT^nv=0$$ Let $m$ be the highest power such that $p_mT^mv$ is nonzero. Let $a$ be some monomial with nonzero coefficient in $p_m$, and consider the Laurent polynomial coefficients of $$a^{-1}(p_mT^mv+\cdots+p_0v)$$ Taking the constant terms is a nontrivial linear combination of $T^mv,T^{m-1}v,\cdots,v$ with coefficients $a_0,a_1,\ldots,a_m$ in $k$ and by homogeneity it must be equal to 0. Specializing the variables can yield any vector in the original vector space. It follows that $a_mT^m+a_{m-1}T^{m-1}+\cdots+a_0I$ is identically zero as a linear transformation, proving the result.

Answer 4

This is an elaboration on roy smith's answer.

For any $x\in V$, we define the local minimal polynomial of $T$ at $x$ as the monic polynomial $m_x$ of the least degree such that $m_x(T)x=0$. Just like the case for the usual minimal polynomial, it can be shown that $m_x$ divides every polynomial $g$ such that $g(T)x=0$.

Let $p_1^{r_1}\cdots p_m^{r_m}$ be a factorisation of the minimal polynomial of $T$ into powers of distinct irreducible factors. Let $q_i=\prod_{j\ne i}p_j^{r_j}$. Note that $V$ is equal to the sum $\ker p_1^{r_1}(T)+\cdots+\ker p_m^{r_m}(T)$, because $\alpha_1q_1+\cdots+\alpha_mq_m=1$ for some polynomials $\alpha_1,\ldots,\alpha_m$ and $\alpha_i(T)q_i(T)x\in\ker p_i^{r_i}(T)$ for every vector $x$. This sum is also a direct sum, for, if $p_i^{r_i}(T)x=0=p_j^{r_j}(T)x$, then $m_x$ divides both $p_i^{r_i}$ and $p_j^{r_j}$. Hence $m_x=1$ and $x=0$.

Knowing that $V$ is a direct sum of those kernels, it remains to show that $\dim\ker p_i^{r_i}(T)\ge\deg\left(p_i^{r_i}\right)$. Suppose $i=1$. Since $p_1^{r_1}\cdots p_m^{r_m}$ is the minimal polynomial of $T$, there exists some $x\in V$ such that $(p_1^{r_1-1}q_1)(T)x\ne0$. Let $y=q_1(T)x$. Then $p_1^{r_1-1}(T)y\ne0=p_1^{r_1}(T)y$. Hence $m_y=p_1^{r_1}$ and by the minimality of $m_y$, $y,\,Ty,\,T^2y,\,\ldots,\,T^{r_1\det(p_1)-1}y$ are linearly independent members of $\ker p_1^{r_1}(T)$. The conclusion now follows.