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Verify that \(\lambda_{i}\) is an eigenvalue of \(A\) and that \(\mathbf{x}_{i}\) is a corresponding eigenvector. $$A=\left[\begin{array}{rrr} 2 & 3 & 1 \\ 0 & -1 & 2 \\ 0 & 0 & 3 \end{array}\right], \begin{array}{l} \lambda_{1}=2, \mathbf{x}_{1}=(1,0,0) \\ \lambda_{2}=-1, \mathbf{x}_{2}=(1,-1,0) \\ \lambda_{3}=3, \mathbf{x}_{3}=(5,1,2) \end{array}$$

Short Answer

Expert verified
Only \(\lambda_{1}=2\) with \(\mathbf{x}_{1}=(1,0,0)\) is an eigenvalue-eigenvector pair. The other pairs are not as they do not satisfy the equation \(A\mathbf{x} = \lambda\mathbf{x}\).

Step by step solution

01

Verification for the first pair \(\lambda_{1}\) and \(\mathbf{x}_{1}\)

For \(\lambda_{1}=2\) with eigenvector \(\mathbf{x}_{1}=(1,0,0)\), you need to apply the equation with the given matrix A. The matrix multiplication for A and \(\mathbf{x}_{1}\) is \(A\mathbf{x}_{1}=[(2*1 + 3*0 + 1*0), (0*1 + -1*0 + 2*0), (0*1 + 0*0 + 3*0)] = [2,0,0]\) which when compared is equal to \(2*\mathbf{x}_{1} = [2,0,0]\). Hence, verifying \(\lambda_{1}\) is an eigenvalue of A and \(\mathbf{x}_{1}\) is a corresponding eigenvector.
02

Verification for the second pair \(\lambda_{2}\) and \(\mathbf{x}_{2}\)

For \(\lambda_{2}=-1\) with eigenvector \(\mathbf{x}_{2}=(1,-1,0)\), similar process as above is performed. The matrix multiplication for A and \(\mathbf{x}_{2}\) is \(A\mathbf{x}_{2}=[(2*1 + 3*-1 + 1*0), (0*1 + -1*-1 + 2*0), (0*1 + 0*-1 + 3*0)] = [2,1,0]\) which when compared is not equal to -1*\(\mathbf{x}_{2} = [-1,1,0]\). Hence, it is observed that \(\lambda_{2}\)=-1 is not an eigenvalue of A and \(\mathbf{x}_{2}\) is not a corresponding eigenvector.
03

Verification for the third pair \(\lambda_{3}\) and \(\mathbf{x}_{3}\)

For \(\lambda_{3}=3\) with eigenvector \(\mathbf{x}_{3}=(5,1,2)\), a similar process is performed. The matrix multiplication for A and \(\mathbf{x}_{3}\) is \(A\mathbf{x}_{3}=[(2*5 + 3*1 + 1*2), (0*5 + -1*1 + 2*2), (0*5 + 0*1 + 3*2)] = [14,-1,6]\) which when compared is not equal to 3*\(\mathbf{x}_{3} = [15,3,6]\). Therefore, it can be concluded that \(\lambda_{3}\)=3 is not an eigenvalue of A and \(\mathbf{x}_{3}\) is not a corresponding eigenvector.

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Key Concepts

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

Matrix Multiplication
At the heart of many linear algebra problems—including those involving eigenvalues and eigenvectors—is matrix multiplication. This fundamental operation is not the same as multiplying individual numbers; instead, it involves a systematic combination of adding and multiplying components of the matrices.

To multiply a matrix by a vector, as in our exercise, we perform a series of dot products between the rows of the matrix and the columns of the vector. The result is a new vector where each component is the sum of products from one row of the matrix and the vector. This is crucial when it comes to verifying whether a given vector is an eigenvector of a matrix, as we need to compare the product of the matrix and the vector to the product of the eigenvalue and the vector.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. One of the key concepts in linear algebra is the idea of eigenvalues and eigenvectors. These are important in understanding the characteristics of linear transformations and are widely used in various fields such as physics, engineering, and computer science.

Linear algebra provides the language and framework for solving systems of equations and transforming spaces. When we consider transformations, eigenvalues tell us how much the eigenvectors are stretched or shrunk, and eigenvectors show us the directions that remain unchanged by these transformations.
Eigenvector Verification
Eigenvector verification is a crucial step in assessing whether a given vector is indeed an eigenvector of a matrix. The process involves two main parts: first, we perform the matrix multiplication of the matrix by the candidate eigenvector. Second, we multiply the candidate eigenvalue by the same vector.

If everything is correct, these two products should be equal. This equality signifies that under the transformation represented by the matrix, the vector is only scaled by the eigenvalue, and its direction remains unaltered. In simpler terms, if the output of the multiplication process is just our initial vector scaled by the eigenvalue, then we have verified that the vector is an eigenvector of the matrix.
Eigenvalue Verification
Eigenvalue verification is the counterpart to verifying an eigenvector. It involves confirming that the eigenvalue associated with a vector is correct. This process takes us back to the equation we use for eigenvectors: for a matrix A and an eigenvalue \( \lambda \), the equation \( A\mathbf{x} = \lambda\mathbf{x} \) must hold for the vector \( \mathbf{x} \) to be an eigenvector.

Determining whether a number is indeed an eigenvalue involves finding out if there's a non-zero vector \( \mathbf{x} \) that satisfies this equation. If our multiplication and comparison show that the two sides of the equation are not equal, the purported eigenvalue is not an actual eigenvalue of the matrix. This step is critical to ensure we have the correct eigenvalues, as these represent the factors by which corresponding eigenvectors are scaled during the transformation.

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Most popular questions from this chapter

A population has the characteristics listed below. (a) A total of \(80 \%\) of the population survives its first year. Of that \(80 \%, 25 \%\) survives the second year. The maximum life span is 3 years. (b) The average number of offspring for each member of the population is 3 the first year, 6 the second year, and 3 the third year. The population now consists of 150 members in each of the three age classes. How many members will there be in each age class in 1 year? In 2 years?

Determine whether the matrix is orthogonal. $$\left[\begin{array}{ccc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} \\ 0 & \frac{\sqrt{6}}{3} & \frac{\sqrt{3}}{3} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} & -\frac{\sqrt{3}}{3} \end{array}\right]$$

Write out the system of first-order linear differential equations represented by the matrix equation \(\mathbf{y}^{\prime}=A \mathbf{y} .\) Then verify the indicated general solution. $$A=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -4 & 0 \end{array}\right], \begin{array}{rr} y_{1}=C_{1}+ & C_{2} \cos 2 t+C_{3} \sin 2 t \\ y_{2}= & 2 C_{3} \cos 2 t-2 C_{2} \sin 2 t \\ y_{3}= & -4 C_{2} \cos 2 t-4 C_{3} \sin 2 t \end{array}$$

Find an orthogonal matrix \(P\) such that \(P^{T} A P\) diagonalizes \(A .\) Verify that \(P^{T} A P\) gives the proper diagonal form. $$A=\left[\begin{array}{rrr} 0 & 10 & 10 \\ 10 & 5 & 0 \\ 10 & 0 & -5 \end{array}\right]$$

Let \(A\) be a diagonalizable \(n \times n\) matrix and \(P\) an invertible \(n \times n\) matrix such that \(B=P^{-1} A P\) is the diagonal form of \(A\) Prove that (a) \(B^{k}=P^{-1} A^{k} P,\) where \(k\) is a positive integer. (b) \(A^{k}=P B^{k} P^{-1},\) where \(k\) is a positive integer.

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