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Chapter 8: Problem 24

Let \(A\) be an \(n \times n\) matrix and \(\mathbf{v}\) an \(n\)-vector such that \(A^{k} \mathbf{v}=\mathbf{0}\) whilst \(A^{k-1} \mathbf{v} \neq 0\). Show that \(\left\\{\mathbf{v}, A \mathbf{v}, A^{2} \mathbf{v}, \ldots, A^{k-1} \mathbf{v}\right\\}\) is an LI set. [Hint: from $$ \mathbf{u}=\alpha_{0} \mathbf{v}+\alpha_{1} A \mathbf{v}+\alpha_{2} A^{2} \mathbf{v}+\ldots+\alpha_{k-1} A^{k-1} \mathbf{v}=0 $$ deduce \(\left.A^{k-1} u=a_{0} A^{k-1} v=\mathbf{0} .\right]\)

Short Answer

Expert verified
The set \(\{\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \ldots, A^{k-1}\mathbf{v}\}\) is linearly independent.

Step by step solution

01

Understand the Problem Statement

We need to show that the set \(\{\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \ldots, A^{k-1}\mathbf{v}\}\) is linearly independent. This means that there are no scalars \(\alpha_0, \alpha_1, \ldots, \alpha_{k-1}\), not all zero, such that the linear combination \(\alpha_0 \mathbf{v} + \alpha_1 A\mathbf{v} + \ldots + \alpha_{k-1} A^{k-1}\mathbf{v} = \mathbf{0}\).
02

Form Linear Combination

Let's assume a linear combination of the set equates to zero: \(\mathbf{u} = \alpha_0 \mathbf{v} + \alpha_1 A\mathbf{v} + \alpha_2 A^2\mathbf{v} + \ldots + \alpha_{k-1} A^{k-1}\mathbf{v} = \mathbf{0}\). Our goal is to derive a contradiction that shows all coefficients must be zero.
03

Apply Matrix \(A\)

Multiply \(\mathbf{u}\) by \(A\) repeatedly until reaching an expression where the last term explicitly involves \(A^{k-1}\mathbf{v}\), which ultimately becomes zero due to \(A^k \mathbf{v} = \mathbf{0}\).
04

Impose Conditions at \(A^{k-1}\mathbf{v}\) and Beyond

Based on our assumption and repeated application of \(A\), at \(A^{k-1}\mathbf{u}\), each term incorporates at least one factor of \(A\) beyond the \(k-1\) applications already used. Since \(A^k\mathbf{v} = \mathbf{0}\), this causes all terms in \(A^{k-1}\mathbf{u}\) to collapse to zero.
05

Show Scalar Dependencies Implies Falsehood

Since \(A^{k-1} \mathbf{u} = \mathbf{0}\), note that \(A^{k-1} \mathbf{u} = \alpha_0 A^{k-1} \mathbf{v}\). Since \(A^{k-1} \mathbf{v} eq \mathbf{0}\), it follows that for the scalar terms to satisfy this dependency, \(\alpha_0 = 0\). Use this to show iteratively that each further term up to \(\alpha_{k-1}\) must also be zero.
06

Conclude with Linear Independence

With all \(\alpha_i = 0\), we conclude that no nontrivial linear combination of the set \(\{\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \ldots, A^{k-1}\mathbf{v}\}\) can be zero, confirming the set is linearly independent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Theory
Matrix Theory is a cornerstone of linear algebra. It involves the study of matrices and their properties, transformations, and operations. In our exercise, we work with a square matrix \(A\), an \(n \times n\) matrix, which entails that it has the same number of rows and columns. This is particularly important when discussing concepts like eigenvectors and transformations because these only apply to square matrices.

In Matrix Theory:
  • A matrix is used to perform linear transformations, which are fundamental in transforming vector spaces.
  • Matrix operations include addition, multiplication, and finding inverses, which have rules similar to those of numerical arithmetic but are more complex.
  • The identity matrix acts as the multiplicative identity in matrix multiplication. An identity matrix has 1s on the diagonal and 0s elsewhere.
Understanding the role and structure of matrices is crucial when analyzing and manipulating transformations between vector spaces.
Vector Spaces
A vector space is a fundamental algebraic structure used in linear algebra. It's a collection of vectors where you can perform vector addition and scalar multiplication and still stay within the space.

A set of vectors is linearly independent if no vector in the set is a linear combination of the others. This is a pivotal concept in vector spaces:
  • Linear independence in our context means that the set \(\{\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \ldots, A^{k-1}\mathbf{v}\}\) cannot have scalars lead to the zero vector unless all scalars are zero.
  • Dimension is essentially the size of the basis of a vector space, which corresponds to the maximum number of linearly independent vectors possible within the space.
  • A basis of a vector space spans it entirely, meaning any vector in the space can be expressed as a combination of basis vectors.
In our exercise, proving linear independence involves ensuring that only the trivial solution (all scalars are zero) leads to the zero vector.
Linear Transformations
Linear transformations are functions between vector spaces that respect the operations of addition and scalar multiplication. They can be represented by matrices, such as the matrix \(A\) in our exercise, which performs a transformation on the vector \(\mathbf{v}\).

Key aspects about linear transformations:
  • Vector transformations: Applying matrix transformations like \(A\mathbf{v}\) moves the vector \(\mathbf{v}\) within the vector space.
  • Matrix properties: The transformations of \(A\) influence the independence and span of vectors, aligning with the properties of the set \(\{\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \ldots, A^{k-1}\mathbf{v}\}\).
  • Kernel and image: The kernel of a transformation is a set of vectors that map to the zero vector, and the image is the set of all outputs.
In our case, showing the set鈥檚 linear independence essentially demonstrates that the transformations do not collapse the vectors into dependency.
Eigenvectors
Eigenvectors are special vectors in Matrix Theory associated with linear transformations. If \(\mathbf{v}\) is an eigenvector of matrix \(A\), then \(A\mathbf{v} = \lambda \mathbf{v}\) for some scalar \(\lambda\), called the eigenvalue.

Understanding eigenvectors:
  • Eigenvectors do not change direction under the transformation, only their magnitude may change, characterized by the eigenvalue \(\lambda\).
  • In the context of linear independence, although our set in the exercise doesn鈥檛 directly use eigenvectors, understanding linear transformations (affected by eigenvectors) is crucial.
  • Determining eigenvectors helps to find bases for invariant subspaces of a matrix, thus playing significant roles in system dynamics and stability analysis.
With the given exercise, recognizing the distinction between eigenvalue problems and linear independence can deepen understanding of vector transformations.

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Most popular questions from this chapter

Is the subset \(\left\\{\left[\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{array}\right],\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\right\\}\) of \(M_{3}(\mathbb{R}) \mathrm{LI} ?\)

Explain why it is not sensible to talk - as one sometimes might be tempted to - of 'a set of linearly (in)dependent vectors'.

Give a geometric argument to show that any set of three vectors in \(\mathbb{R}^{2}\) (any set of four vectors in \(\mathbb{R}^{3}\) ) is LD.

Let \(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\) be a set of non-zero vectors in a vector space \(V\) such that \(\mathbf{v}_{1} \cdot \mathbf{v}_{2}=0\) \(\mathbf{v}_{2} \cdot \mathbf{v}_{3}=0, \mathbf{v}_{3} \cdot \mathbf{v}_{1}=0\). Prove that the set \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is LI. [Hint: assuming that \(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\alpha_{3} \mathbf{v}_{3}=\mathbf{0}\) look at \(\left.\mathbf{v}_{1} \cdot\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\alpha_{3} \mathbf{v}_{3}\right)=\mathbf{v}_{1} \cdot \mathbf{0}=0=\alpha_{1} \mathbf{v}_{1} \cdot \mathbf{v}_{1} \cdot\right]\)

(a) For which scalars \(k\) is \(\\{(1,-1,2),(2,-3, k),(4,2,1)\\}\) an LI set? (b) For which scalars \(k\) is \(\\{(k, 1,0),(1,0, k),(0, k, 1)\\}\) an LI set?
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