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

Solution 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!

Solution 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 $\le \dim(V)$.

Solution 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.