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Chapter 5: Problem 31

Construct a nonzero \(2 \times 2\) matrix that is invertible but not diagonalizable.

Short Answer

Expert verified
A = \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) is an invertible, non-diagonalizable 2x2 matrix.

Step by step solution

01

Understand the Problem

A matrix is invertible if it has a nonzero determinant, and it's diagonalizable if there exists a basis of eigenvectors. We need to find a non-diagonalizable, invertible 2x2 matrix.
02

Identify the Characteristics of the Matrix

For a 2x2 matrix to be invertible, its determinant must be non-zero. A non-diagonalizable matrix with real coefficients usually has a repeated eigenvalue with only one associated linearly independent eigenvector.
03

Choose an Example

We choose the matrix \( A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \). This matrix is a classic example of a non-diagonalizable 2x2 matrix.
04

Verify Invertibility

Calculate the determinant of matrix \( A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \). The determinant is \(1 \times 1 - 0 \times 1 = 1\), which is nonzero. Thus, the matrix is invertible.
05

Check Diagonalizability

Find the eigenvalues of \( A \) by solving \( \det(A - \lambda I) = 0 \). Here, \( \lambda I = \begin{pmatrix} \lambda & 0 \ 0 & \lambda \end{pmatrix} \), and the characteristic equation is \((1-\lambda)^2 = 0\), giving a repeated eigenvalue \( \lambda = 1 \).
06

Confirm Non-diagonalizability

To check diagonalizability, we need two linearly independent eigenvectors for the eigenvalue \( \lambda = 1 \). Solving \( (A - I)\mathbf{v} = 0 \), we find \( \mathbf{v} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \), but only one linearly independent eigenvector exists, confirming that \( A \) is not diagonalizable.

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Key Concepts

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

Invertible Matrix
An invertible matrix, sometimes called a non-singular matrix, is a fundamental concept in linear algebra. To determine if a matrix is invertible, you need to calculate its determinant. Simply put, a matrix is invertible if its determinant is not zero. This means, for a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), it is invertible if \(ad - bc eq 0\).
An invertible matrix has an inverse, denoted as \(A^{-1}\), satisfying the equality \(AA^{-1} = I\), where \(I\) is the identity matrix.
  • The identity matrix is a special type of diagonal matrix where all elements are zero except those on the main diagonal, which are all ones.
  • Having an inverse means you can "reverse" or "undo" the effect of the matrix on vectors.
In context, if you multiply a vector by an invertible matrix and then by the inverse of that matrix, you return to the original vector.
Diagonalizable Matrix
A matrix is considered diagonalizable if it can be expressed in the form \(PDP^{-1}\), where \(D\) is a diagonal matrix composed of the matrix's eigenvalues and \(P\) is a matrix whose columns are formed by the corresponding eigenvectors. Not every matrix is diagonalizable, and there are specific criteria to check if diagonalizability is possible.
For a matrix to be diagonalizable:
  • There must be as many linearly independent eigenvectors as the dimension of the matrix.
  • This often translates to having the complete set of eigenvectors to form the matrix \(P\).
When a matrix is not diagonalizable, like the example matrix \(A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}\), it might have repeated eigenvalues yet lack sufficient eigenvectors because one of the key requirements 鈥� the number of linearly independent eigenvectors 鈥� is not met.
Eigenvectors
Eigenvectors are special vectors associated with a matrix, which remain in the same direction after the matrix is applied to them 鈥� they might only scale in magnitude. If \(A\) is a matrix, and \(\mathbf{v}\) is an eigenvector of \(A\), this relationship is expressed as \(A\mathbf{v} = \lambda \mathbf{v}\), where \(\lambda\) is the eigenvalue associated with that eigenvector.
You can determine eigenvectors by solving the equation \((A - \lambda I)\mathbf{v} = 0\). Important characteristics of eigenvectors include:
  • Eigenvectors are always paired with specific eigenvalues.
  • Linearly independent eigenvectors ensure that a matrix can potentially be diagonalized.
  • In some cases, like repeated eigenvalues, there might be fewer eigenvectors available.
The presence of a complete set of linearly independent eigenvectors is crucial for the possibility of diagonalizing a matrix.
Eigenvalues
Eigenvalues represent scalar values that signify specific features of a matrix. When examining a square matrix \(A\), eigenvalues are the solutions \(\lambda\) to the equation \(\det(A - \lambda I) = 0\). Here, \(I\) is the identity matrix of the same size as \(A\). Each eigenvalue corresponds to one or more eigenvectors.
Characteristics of eigenvalues include:
  • A matrix of size \(n\times n\) can have up to \(n\) eigenvalues.
  • Eigenvalues can be real or complex numbers.
  • If all eigenvalues are distinct, the matrix can be easily diagonalized.
  • Repeated eigenvalues might pose challenges in diagonalization if not enough linearly independent eigenvectors exist.
For example, the matrix \(A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}\) has a repeated eigenvalue of 1, which plays a role in its inability to be diagonalized.

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Most popular questions from this chapter

In Exercises 19 and \(20,\) find \((a)\) the largest eigenvalue and \((b)\) the eigenvalue closest to zero. In each case, set \(\mathbf{x}_{0}=(1,0,0,0)\) and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector. $$ A=\left[\begin{array}{cccc}{10} & {7} & {8} & {7} \\ {7} & {5} & {6} & {5} \\\ {8} & {6} & {10} & {9} \\ {7} & {5} & {9} & {10}\end{array}\right] $$

Let \(A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right] .\) Use formula \((1)\) for a determinant (given before Example 2\()\) to show that det \(A=a d-b c\) Consider two cases: \(a \neq 0\) and \(a=0\)

Exercises \(7-12\) require MATLAB or other computational aid. In Exercises 7 and \(8,\) use the power method with the \(\mathbf{x}_{0}\) given. List \(\left\\{\mathbf{x}_{k}\right\\}\) and \(\left\\{\mu_{k}\right\\}\) for \(k=1, \ldots, 5 .\) In Exercises 9 and \(10,\) list \(\mu_{5}\) and \(\mu_{6} .\) $$ A=\left[\begin{array}{ll}{2} & {1} \\ {4} & {5}\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}{1} \\ {0}\end{array}\right] $$

Another estimate can be made for an eigenvalue when an approximate eigenvector is available. Observe that if \(A \mathbf{x}=\lambda \mathbf{x}\) , then \(\mathbf{x}^{T} A \mathbf{x}=\mathbf{x}^{T}(\lambda \mathbf{x})=\lambda\left(\mathbf{x}^{T} \mathbf{x}\right),\) and the Rayleigh quotient $$R(\mathbf{x})=\frac{\mathbf{X}^{T} A \mathbf{x}}{\mathbf{x}^{T} \mathbf{x}}$$ equals \(\lambda\) . If \(\mathbf{x}\) is close to an eigenvector for \(\lambda,\) then this quotient is close to \(\lambda .\) When \(A\) is a symmetric matrix \(\left(A^{T}=A\right)\) , the Rayleigh quotient \(R\left(\mathbf{x}_{k}\right)=\left(\mathbf{x}_{k}^{T} A \mathbf{x}_{k}\right) /\left(\mathbf{x}_{k}^{T} \mathbf{x}_{k}\right)\) will have roughly twice as many digits of accuracy as the scaling factor \(\mu_{k}\) in the power method. Verify this increased accuracy in Exercises 11 and 12 by computing \(\mu_{k}\) and \(R\left(\mathbf{x}_{k}\right)\) for \(k=1, \ldots, 4\) $$ A=\left[\begin{array}{ll}{5} & {2} \\ {2} & {2}\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}{1} \\ {0}\end{array}\right] $$

Define \(T : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) by \(T(\mathbf{x})=A \mathbf{x},\) where \(A\) is a \(3 \times 3\) matrix with eigenvalues 5 and \(-2 .\) Does there exist a basis \(\mathcal{B}\) for \(\mathbb{R}^{3}\) such that the \(\mathcal{B}\) -matrix for \(T\) is a diagonal matrix? Discuss.
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