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. Anurag Bishnoi says:
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 $T - a_2I$.
1. ayushkhaitan3437 says: