# Minimal polynomials of linear transformations classed as scalar polynomials?

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!

• Both $f_1(T)$ and $f_2(T)$ are polynomials of $T$, and $T$ clearly commutes with (the power of) itself. Jan 27 at 11:59
• 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. Jan 27 at 11:59
• Sorry - replaced the images, should be better now.
– Tom
Jan 27 at 12:09
• Thank you @macton ! Completely forgot about that old trick...
– Tom
Jan 27 at 12:10
• 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. 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.