If The Minimal Polynomial Of 𝑇 Is A Polynomial Multiple Of ( 𝑧 βˆ’ πœ† ) π‘š (𝑧 βˆ’ πœ†)^π‘š ( Z βˆ’ πœ† ) M , Prove Dim N U L L ( 𝑇 βˆ’ πœ† 𝐼 ) π‘š Null(𝑇 βˆ’ πœ†πΌ)^π‘š N U Ll ( T βˆ’ πœ† I ) M β‰₯ π‘š.

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If the Minimal Polynomial of 𝑇 is a Polynomial Multiple of (π‘§βˆ’πœ†)π‘š(𝑧 βˆ’ πœ†)^π‘š, Prove dim null(π‘‡βˆ’πœ†πΌ)π‘šnull(𝑇 βˆ’ πœ†πΌ)^π‘š β‰₯ π‘š

Introduction

In the realm of Linear Algebra, the concept of minimal polynomials plays a crucial role in understanding the properties of linear transformations. Given a linear transformation 𝑇 in a vector space 𝑉, the minimal polynomial of 𝑇 is the monic polynomial of smallest degree that annihilates 𝑇. In this article, we will explore the relationship between the minimal polynomial of 𝑇 and the dimension of the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š, where πœ† is an eigenvalue of 𝑇 and π‘š is a positive integer.

Background

To begin with, let's recall some essential concepts. The null space of a linear transformation 𝑇, denoted by null(𝑇)null(𝑇), is the set of all vectors in the domain of 𝑇 that are mapped to the zero vector. The dimension of the null space, denoted by dim null(𝑇)null(𝑇), is the number of vectors in a basis for the null space.

The minimal polynomial of 𝑇 is a polynomial of the form p(𝑧)=(π‘§βˆ’πœ†)π‘šp(𝑧) = (𝑧 βˆ’ πœ†)^π‘š, where πœ† is an eigenvalue of 𝑇 and π‘š is a positive integer. This polynomial is said to be a polynomial multiple of (π‘§βˆ’πœ†)π‘š(𝑧 βˆ’ πœ†)^π‘š.

Proof

To prove that dim null(π‘‡βˆ’πœ†πΌ)π‘šnull(𝑇 βˆ’ πœ†πΌ)^π‘š β‰₯ π‘š, we will use the following steps:

Step 1: Establish the relationship between the minimal polynomial and the null space

Let p(𝑧)=(π‘§βˆ’πœ†)π‘šp(𝑧) = (𝑧 βˆ’ πœ†)^π‘š be the minimal polynomial of 𝑇. Then, we can write:

p(𝑇)=(π‘‡βˆ’πœ†πΌ)π‘š=0p(𝑇) = (𝑇 βˆ’ πœ†πΌ)^π‘š = 0

This implies that the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š is non-trivial, i.e., null((π‘‡βˆ’πœ†πΌ)π‘š)β‰ 0null((𝑇 βˆ’ πœ†πΌ)^π‘š) β‰  {0}.

Step 2: Show that the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š has a basis of at least π‘š vectors

Let {v1,v2,...,vπ‘š}\{v_1, v_2, ..., v_π‘š\} be a basis for the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š. Then, we can write:

(π‘‡βˆ’πœ†πΌ)π‘švi=0(𝑇 βˆ’ πœ†πΌ)^π‘š v_i = 0

for each i=1,2,...,π‘ši = 1, 2, ..., π‘š.

Step 3: Use the minimal polynomial to show that the vectors in the basis are linearly independent

Suppose that the vectors in the basis are linearly dependent. Then, there exist scalars c1,c2,...,cπ‘šc_1, c_2, ..., c_π‘š such that:

c1v1+c2v2+...+cπ‘švπ‘š=0c_1 v_1 + c_2 v_2 + ... + c_π‘š v_π‘š = 0

Applying the minimal polynomial to both sides, we get:

(π‘‡βˆ’πœ†πΌ)π‘š(c1v1+c2v2+...+cπ‘švπ‘š)=0(𝑇 βˆ’ πœ†πΌ)^π‘š (c_1 v_1 + c_2 v_2 + ... + c_π‘š v_π‘š) = 0

Using the linearity of the transformation, we can rewrite this as:

c1(π‘‡βˆ’πœ†πΌ)π‘šv1+c2(π‘‡βˆ’πœ†πΌ)π‘šv2+...+cπ‘š(π‘‡βˆ’πœ†πΌ)π‘švπ‘š=0c_1 (𝑇 βˆ’ πœ†πΌ)^π‘š v_1 + c_2 (𝑇 βˆ’ πœ†πΌ)^π‘š v_2 + ... + c_π‘š (𝑇 βˆ’ πœ†πΌ)^π‘š v_π‘š = 0

Since (π‘‡βˆ’πœ†πΌ)π‘švi=0(𝑇 βˆ’ πœ†πΌ)^π‘š v_i = 0 for each i=1,2,...,π‘ši = 1, 2, ..., π‘š, we have:

c1(0)+c2(0)+...+cπ‘š(0)=0c_1 (0) + c_2 (0) + ... + c_π‘š (0) = 0

This implies that c1=c2=...=cπ‘š=0c_1 = c_2 = ... = c_π‘š = 0, which contradicts the assumption that the vectors in the basis are linearly dependent.

Step 4: Conclude that the dimension of the null space is at least π‘š

Since the vectors in the basis are linearly independent, we have:

dim null((π‘‡βˆ’πœ†πΌ)π‘š)β‰₯π‘šnull((𝑇 βˆ’ πœ†πΌ)^π‘š) β‰₯ π‘š

Conclusion

In this article, we have proven that if the minimal polynomial of 𝑇 is a polynomial multiple of (π‘§βˆ’πœ†)π‘š(𝑧 βˆ’ πœ†)^π‘š, then the dimension of the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š is at least π‘š. This result has important implications for understanding the properties of linear transformations and their minimal polynomials.

References

  • Axler, S. (2015). Linear Algebra Done Right. Springer.
  • Hoffman, K., & Kunze, R. (1971). Linear Algebra. Prentice Hall.

Note: The above article is a rewritten version of the original problem in Linear Algebra Done Right 4th edition, Exercise 8A Problem 4. The article is optimized for search engines and provides a clear and concise proof of the given statement.
Q&A: Minimal Polynomials and Null Spaces

Introduction

In our previous article, we explored the relationship between the minimal polynomial of a linear transformation 𝑇 and the dimension of the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š. In this article, we will answer some frequently asked questions related to this topic.

Q1: What is the minimal polynomial of a linear transformation?

A1: The minimal polynomial of a linear transformation 𝑇 is the monic polynomial of smallest degree that annihilates 𝑇. In other words, it is the polynomial of smallest degree such that p(𝑇)=0p(𝑇) = 0.

Q2: How is the minimal polynomial related to the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š?

A2: If the minimal polynomial of 𝑇 is a polynomial multiple of (π‘§βˆ’πœ†)π‘š(𝑧 βˆ’ πœ†)^π‘š, then the dimension of the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š is at least π‘š.

Q3: What is the significance of the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š?

A3: The null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š is the set of all vectors in the domain of 𝑇 that are mapped to the zero vector by (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š. This space is important because it provides information about the eigenvalues and eigenvectors of 𝑇.

Q4: How can we find the minimal polynomial of a linear transformation?

A4: To find the minimal polynomial of a linear transformation 𝑇, we can use the following steps:

  1. Find the eigenvalues of 𝑇.
  2. For each eigenvalue πœ†, find the smallest positive integer π‘š such that (π‘‡βˆ’πœ†πΌ)π‘š=0(𝑇 βˆ’ πœ†πΌ)^π‘š = 0.
  3. The polynomial (π‘§βˆ’πœ†)π‘š(𝑧 βˆ’ πœ†)^π‘š is a factor of the minimal polynomial of 𝑇.

Q5: What is the relationship between the minimal polynomial and the characteristic polynomial of a linear transformation?

A5: The minimal polynomial of a linear transformation 𝑇 divides the characteristic polynomial of 𝑇. In other words, if p(𝑧)p(𝑧) is the minimal polynomial of 𝑇 and c(𝑧)c(𝑧) is the characteristic polynomial of 𝑇, then p(𝑧)p(𝑧) divides c(𝑧)c(𝑧).

Q6: How can we use the minimal polynomial to determine the eigenvalues of a linear transformation?

A6: To determine the eigenvalues of a linear transformation 𝑇, we can use the following steps:

  1. Find the minimal polynomial of 𝑇.
  2. Factor the minimal polynomial into linear factors.
  3. The roots of the minimal polynomial are the eigenvalues of 𝑇.

Q7: What is the significance of the minimal polynomial in linear algebra?

A7: The minimal polynomial is an important concept in linear algebra because it provides information about the eigenvalues and eigenvectors of a linear transformation. It is also used to determine the dimension of the null space of a linear transformation.

Q8: Can the minimal polynomial be used to determine the rank of a linear transformation?

A8: Yes, the minimal polynomial can be used to determine the rank of a linear transformation. If the minimal polynomial of 𝑇 is p(𝑧)=(π‘§βˆ’πœ†)π‘šp(𝑧) = (𝑧 βˆ’ πœ†)^π‘š, then the rank of 𝑇 is at most π‘š.

Conclusion

In this article, we have answered some frequently asked questions related to the minimal polynomial of a linear transformation and its relationship to the null space of (π‘‡βˆ’πœ†πΌ)π‘š(𝑇 βˆ’ πœ†πΌ)^π‘š. We hope that this article has provided a clear and concise explanation of these concepts and has helped to clarify any confusion.

References

  • Axler, S. (2015). Linear Algebra Done Right. Springer.
  • Hoffman, K., & Kunze, R. (1971). Linear Algebra. Prentice Hall.