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Chapter 4: Problem 21

The matrices either are not diagonalizable or do not have a dominant eigenvalue (or both). Apply the power method anyway with the given initial vector \(\mathbf{x}_{0}\) performing eight iterations in each case. Compute the exact eigenvalues and eigenvectors and explain what is happening. $$A=\left[\begin{array}{ll} 4 & 1 \\ 0 & 4 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]$$

Short Answer

Expert verified
No dominant eigenvalue; power method lacks convergence to a single eigenvector.

Step by step solution

01

Initialize Power Method

The power method is an iterative process used to estimate the dominant eigenvalue and corresponding eigenvector of a matrix. Start with initial vector \( \mathbf{x}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and matrix \( A = \begin{bmatrix} 4 & 1 \ 0 & 4 \end{bmatrix} \).
02

Perform Iterations

Apply the power method formula \( \mathbf{x}_{k+1} = A \mathbf{x}_k \). After each iteration, normalize the resulting vector to prevent numerical overflow. Repeat this for 8 iterations.
03

First Iteration

Compute \( \mathbf{x}_1 = A \mathbf{x}_0 = \begin{bmatrix} 4 & 1 \ 0 & 4 \end{bmatrix} \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 5 \ 4 \end{bmatrix} \). Normalize to get \( \mathbf{x}_1 = \begin{bmatrix} 0.78 \ 0.62 \end{bmatrix} \).
04

Second Iteration

Compute \( \mathbf{x}_2 = A \mathbf{x}_1 = \begin{bmatrix} 4 & 1 \ 0 & 4 \end{bmatrix} \begin{bmatrix} 0.78 \ 0.62 \end{bmatrix} = \begin{bmatrix} 3.74 \ 2.48 \end{bmatrix} \). Normalize to get \( \mathbf{x}_2 = \begin{bmatrix} 0.84 \ 0.56 \end{bmatrix} \).
05

Third to Eighth Iteration

Continue similarly for further iterations: - \( \mathbf{x}_3 \approx \begin{bmatrix} 0.86 \ 0.51 \end{bmatrix} \)- \( \mathbf{x}_4 \approx \begin{bmatrix} 0.87 \ 0.50 \end{bmatrix} \)- \( \mathbf{x}_5 \approx \begin{bmatrix} 0.88 \ 0.48 \end{bmatrix} \)- \( \mathbf{x}_6 \approx \begin{bmatrix} 0.89 \ 0.46 \end{bmatrix} \)- \( \mathbf{x}_7 \approx \begin{bmatrix} 0.89 \ 0.45 \end{bmatrix} \)- \( \mathbf{x}_8 \approx \begin{bmatrix} 0.90 \ 0.44 \end{bmatrix} \).
06

Compute Exact Eigenvalues and Eigenvectors

Calculate the eigenvalues of \( A \) by solving \( \det(A - \lambda I) = 0 \). This yields \( \lambda_1 = 4 \), \( \lambda_2 = 4 \). These eigenvalues indicate that \( A \) is not diagonalizable and does not have a dominant eigenvalue. Both corresponding eigenvectors are multiples of \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \).
07

Explain the Outcome

The power method is meant to find the dominant eigenvalue, but because \( A \) does not have a true dominant eigenvalue (both are 4), the method does not converge to a meaningful vector related to a single eigenvalue. The procedure results in redundant information as it oscillates in identifying a true dominant eigenvector.

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Key Concepts

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

Eigenvalues
Eigenvalues are fundamental in understanding matrix operations, and they provide insight into the matrix鈥檚 properties and behavior. In simple terms, an eigenvalue is a scalar that scales an eigenvector when a matrix is multiplied by the eigenvector. For the matrix given in the exercise, we solve the characteristic polynomial \( \det(A - \lambda I) = 0 \) to find the eigenvalues. Upon solving, this yields the repeated eigenvalue \( \lambda = 4 \).
  • The equality \( A\mathbf{v} = \lambda\mathbf{v} \) must hold for \( \mathbf{v} \), a non-zero vector.
  • Eigenvalues are intrinsic to the matrix and do not change with scalar multiplication or addition of a constant to the matrix.
In matrices with repeated eigenvalues, such as our case, the system can have "degenerate" behavior, which sometimes complicates finding unique eigenvectors.
Eigenvectors
Eigenvectors are vectors that change only in scale, not direction, when a matrix transformation is applied. Think of them as the "directions" that remain unchanged no matter how we transform our space using the matrix operations that involve the specified matrix. For a matrix \( A \), if \( \lambda \) is an eigenvalue, then any vector \( \mathbf{v} \) that satisfies \( A\mathbf{v} = \lambda \mathbf{v} \) is an eigenvector.
  • Eigenvectors are found by solving the system \((A - \lambda I)\mathbf{v} = 0\).
  • In this exercise, the matrix \( A \) produces eigenvectors that are parallel to \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \), corresponding to the eigenvalue \( 4 \).
This multiplicity (both eigenvectors corresponding to the same eigenvalue) plays a crucial role in whether the matrix can be diagonalized鈥攈ere, it cannot.
Non-diagonalizable Matrix
In linear algebra, a matrix is non-diagonalizable if it cannot be expressed in the form \( PDP^{-1} \) for some diagonal matrix \( D \), where \( P \) is an invertible matrix whose columns are eigenvectors of the matrix. The matrix from the exercise, \( A \), is a classic example of this.
  • Because it has repeated eigenvalues without a full set of linearly independent eigenvectors, it cannot be diagonalized.
  • This nature affects computational techniques such as the Power Method, as seen in the exercise. Without distinct dominant eigenvalues, convergence to meaningful vectors is not achievable.
A non-diagonalizable matrix can still be represented using Jordan canonical form, but this advanced topic goes beyond the scope of this exercise.
Iterative Process
The iterative process is integral to the Power Method, a technique used to approximate the dominant eigenvalue and its corresponding eigenvector. Despite \( A \) being non-diagonalizable, we attempt this process, initializing with an arbitrary vector and repeatedly applying the matrix \( A \) over several iterations.
  • The primary aim is to amplify the direction of the dominant eigenvector within each iteration step.
  • Normalization is crucial to avoid numerical overflow and helps in focusing the results on direction rather than magnitude.
Here, after eight iterations, due to the absence of a distinct dominant eigenvalue, the Power Method could not converge to a significant result. This showcases an essential limitation of the method in handling non-diagonalizable or evenly dominant eigenvalue cases.

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Most popular questions from this chapter

Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of t.) $$\begin{array}{l} x^{\prime}=x+3 y, \quad x(0)=0 \\ y^{\prime}=2 x+2 y, \quad y(0)=5 \end{array}$$

Calculate the positive eigenvalue and \(a\) corresponding positive eigenvector of the Leslie matrix \(L\). $$L=\left[\begin{array}{ll} 1 & 1.5 \\ 0.5 & 0 \end{array}\right]$$

Consider the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\). (a) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) (b) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) (c) Using eigenvalues and eigenvectors, classify the origin as an attractor, repeller, saddle point, or none of these. (d) Sketch several typical trajectories of the system. $$A=\left[\begin{array}{lr} 0 & -1.5 \\ 1.2 & 3.6 \end{array}\right]$$

(a) Use mathematical induction to prove that, for \(n \geq 2,\) the companion matrix \(C(p)\) of \(p(x)=x^{n}+\) \(a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) has characteristic polynomial \((-1)^{n} p(\lambda) .\) [Hint: Expand by cofactors along the last column. You may find it helpful to introduce the polynomial \(q(x)=\left(p(x)-a_{0}\right) / x .\) (b) Show that if \(\lambda\) is an eigenvalue of the companion matrix \(C(p)\) in Equation \((4),\) then an eigenvector corresponding to \(\lambda\) is given by $$\left[\begin{array}{c} \lambda^{n-1} \\ \lambda^{n-2} \\ \vdots \\ \lambda \\ 1 \end{array}\right]$$ If \(p(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) and \(A\) is a square matrix, we can define a square matrix \(p(A) b y\) $$p(A)=A^{n}+a_{n-1} A^{n-1}+\cdots+a_{1} A+a_{0} I$$ An important theorem in advanced linear algebra says that if \(c_{A}(\lambda)\) is the characteristic polynomial of the matrix \(A\), then \(c_{A}(A)=O\) (in words, every matrix satisfies its characteristic equation). This is the celebrated Cayley-Hamilton Theorem, named after Arthur Cayley \((1821-1895)\) pictured below, and sir William Rowan Hamilton (see page 2 ). Cayley proved this theorem in \(1858 .\) Hamilton discovered it, independently, in his work on quaternions, a generalization of the complex numbers.

(a) Give an example to show that if \(A\) can be partitioned as \\[A=\left[\begin{array}{ll}P & Q \\\R & S\end{array}\right]\\] where \(P, Q, R,\) and \(S\) are all square, then it is not necessarily true that \\[\operatorname{det} A=(\operatorname{det} P)(\operatorname{det} S)-(\operatorname{det} Q)(\operatorname{det} R)\\] (b) Assume that \(A\) is partitioned as in part (a) and that \(P\) is invertible. Let \\[B=\left[\begin{array}{cc}P^{-1} & O \\\\-R P^{-1} & I\end{array}\right]\\] Compute det \((B A)\) using Exercise 69 and use the result to show that \\[\operatorname{det} A=\operatorname{det} P \operatorname{det}\left(S-R P^{-1} Q\right)\\] [The matrix \(S-R P^{-1} Q\) is called the Schur complement of \(P\) in \(A\), after Issai Schur \((1875-1941)\) who was born in Belarus but spent most of his life in Germany. He is known mainly for his fundamental work on the representation theory of groups, but he also worked in number theory, analysis, and other areas. (c) Assume that \(A\) is partitioned as in part (a), that \(P\) is invertible, and that \(P R=R P\). Prove that \\[\operatorname{det} A=\operatorname{det}(P S-R Q)\\]
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