In Linear algebra, the characteristic polynomial and the minimal polynomial are the two most essential polynomials that are strongly related to the linear transformation in the n-dimensional vector space V. In this article, we will learn the definition and theorems of a minimal polynomial, as well as several solved examples.

Table of Contents:

• Definition
• Theorem
• Proof
• Examples
• FAQs

Minimal Polynomial Definition

As we know, a monic polynomial is defined as a polynomial whose highest degree coefficient is equal to 1. The definition of the minimal polynomial is based on the monic polynomial. Now, let us look at the minimal polynomial definition.

Definition:

On a Finite Dimensional Vector Space (FDVS), assume that T is a linear operator. If p(t) is a monic polynomial of least positive degree for which p(T) = 0, i.e. the zero operator, then the polynomial p(t) is called a minimal polynomial of T.

Minimal Polynomial Theorem

Assume that p(t) is a minimal polynomial of a linear operator T on a Finite Dimensional Vector Space V.

1. If g(T) = 0, then p(t) divides g(t), for any polynomial g(t). In specific, the minimal polynomial p(t) divides the characteristic polynomial of T.
2. T’s minimal polynomial is unique

Minimal Polynomial Proof

(1): Let us consider g(t) is a polynomial, in which g(T) = 0.

Using the division algorithm, there exist polynomials, say, q(t) and r(t) such that

g(t) = q(t) p(t) + r(t)

where r(t) = 0 or deg r(t) < deg p(t).

Now, we can write

g(T) = q(T) p(T) + r(T)

i.e. 0 = q(T). 0 + r(T)

It means that r(T) = 0.

Since deg r(t) < deg p(t) and p(t) is considered to be the minimal polynomial of T.

Thus, r(t) should be zero.

Therefore , p(t) divides g(t).

Hence, proved.

(2):

Assume that p₁(t) and p₂(t) are both T’s minimal polynomials. p₁(t) then divides p₂(t) by part (1). We have p₂(t) = c p₁(t) for some nonzero scalar c since p₁(t) and p₂(t) have the same degree. And c = 1 since p₁(t) and p₂(t) are monic. As a result, p₁(t) = p₂(t).

Solved Examples on Minimal Polynomial

Example 1:

Find the minimal Polynomial of the matrix:

Solution:

As we know that the characteristic polynomial of A is det(A – tI).

Hence,

I.e., f(t) = – (t – 2)²(t – 3).

Since the minimal polynomial p(t) divides f(t), they should have the same zeros

Hence, the possibilities for minimal polynomial, p(t) are

(t – 2)(t – 3) or (t – 2)²(t – 3)

If suppose, p(t) = (t – 2)(t – 3) then p(A) becomes

P(A) = (A – 2I)(A – 3I)

On solving the above equation, we get

P(A) = A² – 5A + 6I = 0.

Hence, p(t) is a polynomial of least degree, which satisfies p(A) = 0.

Therefore the minimal polynomial of a given matrix A is p(t) = (t – 2)(t – 3).

Example 2:

Suppose T be the linear operator on R² defined by T(a, b) = ( 2a+5b, 6a+b ). Find the minimal polynomial of T.

Solution:

Assume that β be the standard ordered basis for R² .

So,

Hence, The characteristic polynomial of T is given by

f(t) = (t-7) + (t+4)

Since the characteristic polynomial and minimal polynomial have the same zeros, we can conclude that the minimal polynomial is also (t – 7) (t + 4).