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I am trying to learn Minimal Polynomial. I come across a topic seems like the combination of Minimal Polynomial with differentiation operator. However, I don't understand why this lecture note (p.25~p.26) follow this.

$V := \{ f \in K[X] | deg(f) < 4$, and $T$ is the differentiation operator $\sum a_k X^k \to \sum k a_k X^{k−1} \} $

Step 1: Choose obvious basis $E := \{1,X,X^2,X^3 \}$ of $V$

Step 2: Work out the matrix $A$ of the transform/differentiation operator $T$ with respect to basis $E$. $T(1)=0 \to (0,0,0,0)^T$ ;

$T(X)=1 \to (1,0,0,0)^T$ ;

$T(X^2)=2X \to (0,2,0,0)^T$ ;

$T(X^3)=3X^2 \to (0,0,3,0)^T$.

Hence,

$A=\begin{bmatrix}0&1&0&0\\0&0&2&0 \\ 0&0&0&3 \\ 0&0&0&0\end{bmatrix}$

Step 3: Find the characteristic polynomial. $\chi(\lambda)=\lambda^4$.

Step 4: By using Cayley–Hamilton theorem, there are four candidates for minimal polynomial:

(1) $\lambda$ Check (A). Since $A \neq 0$ so it's not minimal polynomial

(2) $\lambda^2$ Check (A)(A). Since $A^2 \neq 0$ so it's not minimal polynomial

(3) $\lambda^3$ Check (A)(A)(A). Since $A^3 \neq 0$ so it's not minimal polynomial

(4) $\lambda^4$ Check (A)(A)(A)(A). Since $A^4=0$ so $\lambda^4$ is the minimal polynomial.

My question is :

  1. In step 1, when we look at differentiation operator, we always look at basis like this? $E := \{1,X,X^2,X^3 \}$

  2. What's the difference between "given a matrix and find its minimal polynomial" and "given a operator and find it minimal polynomial"? The former one has already generate the matrix for us, so the process will reduce to step 3&4? How about the latter one? We will have to "imagine" a matrix which tell us what is this operator going to do?

Thank you!

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In 1, the answer is no. In general, you can freely pick a basis but most of the times in mathematics what you want is to simplify your calculations and the usual basis do this job.

In 2, it depends on how the minimal polynomial was defined. Usually it's defined for linear operators and then extended for matrices, because the conditions you're working on: Given a field $F$, a $F$-vector space $V$ and an operator $T:V\to V$, the set $$\{f(x)\in F[X]:f(T)=0\}$$ it's a non-empty ideal (verify!) of the polynomial ring $F[X]$, which is also a PID (Principal Ideal Domain), so it must be of the form $(g(x))=\{g(x)q(x):q(x)\in F[X]\}$, i.e., generated by some element. The minimal polynomial is then defined as the monic polynomial of least degree that is a generator of this set. For this reason is that that the minimal polynomial is the polynomial of least degree which annihilates the operator $T$.

From this perspective, it seems like matrices are off the table. Nevertheless, knowing that choosing a basis $\mathcal{B}$ produces an unique matrix representation of your operator, all the above can then be translated to matrix language. So if you are given a matrix, you don't need to imagine your operator, you can in fact know how is your operator acting, simply by fixing your field,your vector space and your basis.

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For polynomials of degree at most 3, the basis $\{1,x,x^2,x^3\}$ is particularly convenient because you can see that the matrix corresponding to the differentiation operator with respect to this basis is strictly upper diagonal; this made finding the characteristic polynomial, and thus minimal polynomial, particularly simple.

In general, you can think about matrices as being just a particular way of encoding all the information about a finite-dimensional linear operator, which has a well-defined action on all elements of the space. But to get it in such a compact form, we need to choose a particular basis. Some bases will yield a simpler form of the operator than others; this is an example.

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