TITLE: Minimal polynominal: geometric meaning
QUESTION [11 upvotes]: I am currently studying Chapter 6 of Hoffman & Kunze's Linear Algebra which deals with characteristic values and triangulation and diagonalization theorems. 
The chapter makes heavy use of the concept of the minimal polynomial which it defines as the monic polynomial of the smallest degree that annihilates a linear transformation. 
I am currently finding the proofs in the book that use this definition of the minimal polynomial to be very opaque: just chains of calculations with polynomials that turn out to give the right answer in the end.
I was therefore wondering if there is a more geometric view of the minimal polynomial? What information about the linear transformation is it carrying and why? I guess the answer would be simpler in the case of an algebraically closed field but is there an answer which also works for a general field?

REPLY [5 votes]: Consider the following matrices: 
$$
A = \left(\begin{array}{cc}2&0\\0&2\end{array}\right) \ \ \text{ and } \ \ 
B = \left(\begin{array}{cc}2&1\\0&2\end{array}\right).
$$
The first matrix has minimal polynomial $X - 2$ and the second has minimal polynomial $(X-2)^2$. If we subtract $2I_2$ from these matrices then we get 
$$
\left(\begin{array}{cc}0&0\\0&0\end{array}\right) \ \ \text{ and } \ \ 
\left(\begin{array}{cc}0&1\\0&0\end{array}\right), 
$$
where the first has minimal polynomial $X$ and the second has minimal polynomial $X^2$. 
The different exponents here reflect the fact that a matrix can have a power that is $O$ without being $O$ itself (this doesn't happen with ordinary numbers, like in ${\mathbf R}$ or ${\mathbf C}$). A matrix has a power equal to $O$ precisely when its minimal polynomial is some power of $X$, and the exponent you need on $X$ to achieve that can vary. As another example, compare
$$
M = 
\left(\begin{array}{ccc}
0 & 1 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)
\ \ \text{ and } \ \ 
N = 
\left(\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right).
$$
These are not the zero matrix, but $N^2 = O$ while $M^3 = O$ and $M^2 \not= O$. So $M$ has minimal polynomial $X^3$ and $N$ has minimal polynomial $X^2$.
To describe a (monic) polynomial we could provide its roots and the multiplicities for those roots. For a square matrix, the roots of its minimal polynomial are easy to connect with the matrix: they are the eigenvalues of the matrix. The subtle part is their multiplicities, which are more algebraic than geometric.
It might be natural to hope that the multiplicity of an eigenvalue $\lambda$ as a root of the minimal polynomial is the dimension of the $\lambda$-eigenspace, but this is false in general, as we can see with the matrices $A$ and $B$ above (e.g., $B$ has minimal polynomial $(X-2)^2$ but its 2-eigenspace is 1-dimensional). In the case when the matrix has all distinct eigenvalues, the minimal polynomial is the characteristic polynomial, so you could think of the distinction between the minimal and characteristic polynomials as reflecting the presence of repeated eigenvalues.