I'm currently trying to cover minimal polynomials of linear transformations and their properties, but have become quite confused. I'm all over the basics of matrices, vector spaces, linear transformations and eigenvalues/vectors. The definition given in the notes I'm following is:

"Definition: Let $T:V\rightarrow V$ be a linear transformation from the $\mathbb{F}$-vector space V to itself. Let $m_T$ be the monic polynomial of least degree in the set of polynomials $\{0 \neq g(x)\in\mathbb{F}[x]|g(T)=0\}$. Call $m_T$ the minimal polynomial of T."

What confuses me is the fact that $g(x)\in\mathbb{F}[x]$, however $T\notin\mathbb{F}$; so why are we talking about g(T)? Shouldn't g be acting over scalars, not matrices?

I was ready to just accept this and then move on, but then later I am encountering situations where multiplication of functions taking T is considered commutative. For instance, an excerpt from the notes:

"The polynomial $f(x)= f_1(x)f_2(x)$ satisfies

$f(T)(v_1+v_2) = f(T)(v_1) + f(T)(v_2)$

$= f_2(T)f_1(T)(v_1) + f_1(T)f_2(T)(v_2)$"

T is in essence a matrix here, so why is multiplication of functions of T considered commutative? This seems especially blasphemous to me. It seems it all comes back to the classification of $f(x)\in\mathbb{F}[x]$, which appears to be the root of the issue for me.

Any guidance would be greatly appreciated. Thank you!

  • 1
    $\begingroup$ Both $f_1(T)$ and $f_2(T)$ are polynomials of $T$, and $T$ clearly commutes with (the power of) itself. $\endgroup$
    – macton
    Jan 27 at 11:59
  • $\begingroup$ Please replace the images of text by actual text with formula typeset using MathJax. Images are not accessible for users with screen readers and can't be indexed by search engines. $\endgroup$
    – Christoph
    Jan 27 at 11:59
  • $\begingroup$ Sorry - replaced the images, should be better now. $\endgroup$
    – Tom
    Jan 27 at 12:09
  • $\begingroup$ Thank you @macton ! Completely forgot about that old trick... $\endgroup$
    – Tom
    Jan 27 at 12:10
  • $\begingroup$ The point is that if $T$ is linear, then so is $T^2$, $T^3$, ..., and all linear combinations thereof. So for any polynomial $f$, the value $f(T)$ is a linear operator, and it makes sense to ask whether it is the zero operator. $\endgroup$
    – Joppy
    Jan 27 at 12:34

It's all about notation. Given a polynomial $g(x)=\sum_{i=0}^{n}a_ix^i\in\mathbb{F}[x]$, when you write $g(\lambda)$ where $\lambda\in\mathbb{F}$ you're referring to $\sum_{i=0}^{n}a_i\lambda^i\in\mathbb{F}$, which is a scalar because it's the sum and product of scalars. However, given a $\mathbb{F}$-linear map $T:V\rightarrow V$, $g(T)$ denotes a different object: the $\mathbb{F}$-linear map $\sum_{i=0}^{n}a_iT^i$, where $T^i=T\circ\overset{i)}{\dots}\circ T$ is the composition of functions.


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