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Chapter 7: Problem 12

Illustrate with a simple example that an \(n\)-rowed square matrix may not have eigenvectors which constitute a basis for \(\mathbf{R}^{n}\) (or \(\left.\mathbf{C}^{\prime \prime}\right)\). For instance, consider $$ A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] $$

Short Answer

Expert verified
Matrix A lacks enough eigenvectors to form a basis for \( \mathbf{R}^2 \).

Step by step solution

01

Understanding Eigenvectors and Eigenvalues

An eigenvector of a matrix \( A \) is a non-zero vector \( \mathbf{v} \) such that \( A\mathbf{v} = \lambda \mathbf{v} \), where \( \lambda \) is the corresponding eigenvalue. For a matrix to have a complete eigenbasis for \( \mathbf{R}^n \), it must have \( n \) linearly independent eigenvectors.
02

Calculate Eigenvalues of Matrix A

Given \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \), we find the eigenvalues by solving \( \det(A - \lambda I) = 0 \). This becomes \( \det\left(\begin{bmatrix} 1-\lambda & 1 \ 0 & 1-\lambda \end{bmatrix}\right) = (1-\lambda)^2 = 0 \). The eigenvalue is \( \lambda = 1 \).
03

Find Eigenvectors for the Eigenvalue

Substitute \( \lambda = 1 \) into \( A - \lambda I = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \), and solve \( (A - I)\mathbf{v} = \mathbf{0} \). This gives the system \( x_2 = 0 \), so the eigenvector is \( \mathbf{v} = \begin{bmatrix} 1 \ 0 \end{bmatrix}k \) for any \( k eq 0 \). Thus, the only linearly independent eigenvector is \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \).
04

Analyze Basis Formed by Eigenvectors

Since matrix \( A \) is 2x2, we need two linearly independent eigenvectors to form a complete basis for \( \mathbf{R}^2 \). Here, we only have one eigenvector \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \), which is not enough to form a basis for \( \mathbf{R}^2 \).
05

Conclusion

Matrix \( A \) has repeated eigenvalues but does not have enough eigenvectors to form a basis for \( \mathbf{R}^2 \). This example illustrates that not all square matrices have enough eigenvectors to form a basis for their vector space.

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Key Concepts

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

Eigenvalues
Understanding eigenvalues is crucial when studying the characteristics of matrices. An eigenvalue of a matrix \( A \) is a scalar \( \lambda \) such that there exists a non-zero vector \( \mathbf{v} \) where the equation \( A\mathbf{v} = \lambda \mathbf{v} \) holds true.
To find an eigenvalue of a matrix, one must solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. This determinant equation gives the values of \( \lambda \) for which the matrix \( A \) minus \( \lambda \) times the identity matrix has a zero determinant. This means the matrix \( (A - \lambda I) \) is singular, and its nullspace contains the eigenvectors.
For example, consider matrix \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \). Solving the equation \( \det(A - \lambda I) = 0 \) results in \( (1 - \lambda)^2 = 0 \), giving us the eigenvalue \( \lambda = 1 \). This indicates the presence of repeated eigenvalues, which can sometimes lead to complications in diagonalization and finding a sufficient number of eigenvectors.
Matrix Basis
The concept of a matrix basis is linked to the set of vectors that can span a vector space. For an \( n \)-dimensional vector space like \( \mathbf{R}^n \), the basis should consist of \( n \) linearly independent vectors. These vectors provide a framework upon which any vector in the space can be expressed as a linear combination.
In linear algebra, finding a complete set of eigenvectors for a matrix allows these vectors to form a basis for the vector space. This is known as an eigenbasis. However, not every square matrix can produce enough linearly independent eigenvectors to complete a basis.
With our given matrix \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \), there is only one linearly independent eigenvector, meaning we cannot form a complete basis for \( \mathbf{R}^2 \). For this reason, \( A \) demonstrates that having repeated eigenvalues can lead to insufficient eigenvectors to span the entire space.
Linear Independence
Linear independence is a fundamental concept when discussing vector spaces and basis. A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the others. This means that the determinant of the matrix composed of these vectors is non-zero.
For a matrix to have an eigenbasis, it must produce enough linearly independent eigenvectors. An \( n \times n \) matrix requires \( n \) linearly independent eigenvectors. However, if the matrix has repeated eigenvalues, it may not yield the necessary number of eigenvectors.
In our exercise with matrix \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \), we found only one eigenvector, \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \), which is linearly independent by itself but not sufficient to span \( \mathbf{R}^2 \). This serves as an example of how a matrix may have its eigenvectors form a linearly dependent set, preventing it from creating a full basis for the vector space.

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Most popular questions from this chapter

Show that a two-rowed nonsingular matrix $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \quad \text { has the inverse } \quad A^{-1}=\frac{1}{\operatorname{det} A}\left[\begin{array}{rr} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{array}\right] $$ How does it follow from this formula that \(A^{-1}\) has the eigenvalues \(1 / \lambda_{1}, 1 / \lambda_{2}\), where \(\lambda_{1}, \lambda_{2}\) are the eigenvalues of \(A\) ?

The behavior of the various parts of the spectrum under extension of an operator is of practical interest. If \(T\) is a bounded linear operator and \(T_{1}\) is a linear extension of \(T\), show that we have \(\sigma_{p}\left(T_{1}\right) \supset \sigma_{p}(T)\) and for any \(\lambda \in \sigma_{p}(T)\) the eigenspace of \(T\) is contained in the eigenspace of \(T_{1}\).

A subset \(A_{1}\) of an algebra \(A\) is called a subalgebra of \(A\) if the application of the algebraic operations to elements of \(A_{1}\) yields again elements of \(A_{1}\). The center \(C\) of \(A\) is the set of those elements of \(A\) which commute with all elements of \(A\). Give examples. Show that \(C\) is a commutative subalgebra of \(A\).

If \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) are Cauchy sequences in a normed algebra \(A\), show that \(\left(x_{n} y_{n}\right)\) is Cauchy in \(A\). Moreover, if \(x_{n} \longrightarrow x\) and \(y_{n} \longrightarrow y\), show that \(x_{n} y_{n} \longrightarrow x y\).

Find the eigenvalues and eigenvectors of the following matrices, where \(a\) and \(b\) are real and \(b \neq 0\). $$ A=\left[\begin{array}{rr} 1 & 2 \\ -8 & 11 \end{array}\right], \quad B=\left[\begin{array}{rr} a & b \\ -b & a \end{array}\right] $$
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