# Minimal Polynomials of Linear transformations

I’m prepared to embarrass myself by writing about something that should have been clear to me a long time ago.

This is regarding something about the minimal polynomials of linear transformations that has always confused me. Let $T\in L(V,V)$, where $V$ is an $n$-dimensional vector space. Let us also assume that $T$ has $\{v_1,v_2,\dots,v_n\}$ as distinct eigenvectors, but the corresponding $n$ eigenvalues may not be distinct. If the eigenvalues are $\{a_1,a_2,\dots,a_k\}$ where $k\leq n$, it is then known that $(x-a_1)(x-a_2)\dots (x-a_k)$ is the minimal polynomial of $T$.

We know that as polynomials, $(x-a_1)(x-a_2)\dots (x-a_k)$ and $(x-a_2)(x-a_1)\dots (x-a_k)$ are the same (note that I’ve exchanged the places of $a_1$ and $a_2$. However, when we substitute $x=T$, are $(T-a_1I)(T-a_2I)\dots (T-a_kI)$ and $(T-a_2T)(T-a_1T)\dots (T-a_kT)$ also the same? Remember that matrices are in general not commutative. In fact, if for matrices $A$ and $B$ we have $AB=0$, then it is not necessary that $BA=0$ too.

An earlier exercise in the book “Linear Algebra” by Curtis says that for $f,g\in F[x]$, $f(T)g(T)=g(T)f(T)$. Why is this? Because we’re ultimately going to get the same polynomial in terms of $T$. My mental block came from the fact that I was imagining $T-a_iI$ to be a matrix which I didn’t know much about. I forgot that $T-a_iI$ is a decomposition of a single matrix into two, and matrix multiplication, like the multiplication of complex numbers, is distributive. Hence everything works out as planned.

1. In the last line of second paragraph, “If the eigenvectors …” you mean eigenvalues and not eigenvectors. There is another typo where you write $T - a_2T$ instead of $T - a_2I$.