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Chapter 7: Problem 13

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rr}1 & 6 \\\2 & -3\end{array}\right]$$.

Short Answer

Expert verified
The eigenvalues and corresponding eigenvectors for the given matrix are: Eigenvalue \(位_1 = 1 + 3\sqrt{5}\) with eigenvector \(饾懃_1 = \begin{bmatrix} 1 \\ \frac{1}{2} + \frac{1}{2}\sqrt{5} \end{bmatrix}\), and Eigenvalue \(位_2 = 1 - 3\sqrt{5}\) with eigenvector \(饾懃_2 = \begin{bmatrix} 1 \\ \frac{1}{2} - \frac{1}{2}\sqrt{5} \end{bmatrix}\).

Step by step solution

01

Calculate the Characteristic Equation

To find the eigenvalues, we first need to determine the characteristic equation of the matrix. The characteristic equation is given by: \( det(A - 位I) = 0 \) where A is the given matrix, 位 is the eigenvalue, and I is the identity matrix. The given matrix A is: $$A=\left[\begin{array}{rr}1 & 6 \\\2 & -3\end{array}\right]$$ Let's form the matrix (A - 位I): \((A - 位I) = \begin{bmatrix} 1 - 位 & 6 \\ 2 & -3 - 位 \end{bmatrix}\) Now, we can find the determinant of this matrix and solve it equal to 0.
02

Solve the Characteristic Equation for 位

The determinant of (A - 位I) is given by: \(det(A - 位I) = (1 - 位)(-3 - 位) - (6)(2) \) Solve for 位: \((1 - 位)(-3 - 位) - 12 = 0\) Expanding and simplifying the equation: \(位^2 + 2位 - 9 = 0\) Now we need to find the roots of this quadratic equation. These roots will be the eigenvalues of the given matrix A.
03

Calculate the Eigenvalues

We can find the roots of the quadratic equation \(位^2 + 2位 - 9 = 0\) by using the quadratic formula: \(位 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where a = 1, b = 2, and c = -9. Plugging in these values: \(位 = \frac{-2 \pm \sqrt{2^2 - 4(1)(-9)}}{2(1)}\) \(位 = \frac{-2 \pm \sqrt{40}}{2}\) This gives us two eigenvalues: \(位_1 = \frac{-2 + \sqrt{40}}{2} = 1 + 3\sqrt{5} \) and \(位_2 = \frac{-2 - \sqrt{40}}{2} = 1 - 3\sqrt{5} \)
04

Calculate the Eigenvectors

For each eigenvalue, we need to find the corresponding eigenvectors by solving the equation: \((A - 位I)饾懃 = 0\) For \(位_1 = 1 + 3\sqrt{5}\): \((A - 位_1I) = \begin{bmatrix} -3\sqrt{5} & 6 \\ 2 & -4-3\sqrt{5} \end{bmatrix}\) The augmented matrix for this system is: \(\left[\begin{array}{cc|c} -3\sqrt{5} & 6 & 0 \\ 2 & -4-3\sqrt{5} & 0 \end{array}\right]\) We can solve this system using elementary row operations to find the eigenvector corresponding to 位鈧�. In this case, the eigenvector is: \(饾懃_1 = \begin{bmatrix} 1 \\ \frac{1}{2} + \frac{1}{2}\sqrt{5} \end{bmatrix}\) For \(位_2 = 1 - 3\sqrt{5}\): \((A - 位_2I) = \begin{bmatrix} 3\sqrt{5} & 6 \\ 2 & -4+3\sqrt{5} \end{bmatrix}\) The augmented matrix for this system is: \(\left[\begin{array}{cc|c} 3\sqrt{5} & 6 & 0 \\ 2 & -4+3\sqrt{5} & 0 \end{array}\right]\) We can solve this system using elementary row operations to find the eigenvector corresponding to 位鈧�. In this case, the eigenvector is: \(饾懃_2 = \begin{bmatrix} 1 \\ \frac{1}{2} - \frac{1}{2}\sqrt{5} \end{bmatrix}\)
05

Final Answer

The eigenvalues and their corresponding eigenvectors for the given matrix A are: Eigenvalue \(位_1 = 1 + 3\sqrt{5}\) with eigenvector \(饾懃_1 = \begin{bmatrix} 1 \\ \frac{1}{2} + \frac{1}{2}\sqrt{5} \end{bmatrix}\), and Eigenvalue \(位_2 = 1 - 3\sqrt{5}\) with eigenvector \(饾懃_2 = \begin{bmatrix} 1 \\ \frac{1}{2} - \frac{1}{2}\sqrt{5} \end{bmatrix}\).

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Key Concepts

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

Characteristic Equation
To determine the eigenvalues of a matrix, we employ the characteristic equation. The characteristic equation is a polynomial equation derived from the matrix, which helps us identify these eigenvalues. We start by using the formula:
  • Characteristic Equation: \( \det(A - \lambda I) = 0 \)
Here, \( A \) is the matrix in question, \( \lambda \) symbolizes the eigenvalue, and \( I \) stands for the identity matrix of the same size as \( A \). The process begins by substituting \( \lambda \) into the matrix \( A \) to create \( (A - \lambda I) \). This new matrix is then used to calculate the determinant, leading us to the characteristic equation. Solving this equation for \( \lambda \) reveals the potential eigenvalues of matrix \( A \). Through the roots of the characteristic equation, we come closer to understanding the inherent "directions" and "magnitudes" via eigenvectors connected to a matrix.
Determinant
A determinant is a special number associated with a square matrix. It plays a crucial role in a variety of matrix-related applications, including the determination of eigenvalues. When calculating eigenvalues, the determinant of \( (A - \lambda I) \) is pivotal.
  • Determinant Expression: \( \det(A - \lambda I) \)
This determinant helps to find the characteristic polynomial. Calculating the determinant involves performing specific arithmetic operations on the elements of the matrix. For a 2x2 matrix like \[\begin{bmatrix} a & b \ c & d \end{bmatrix},\]the determinant is given by:\[ \det = ad - bc \]Once calculated, this determinant is set to zero when solving for eigenvalues. It's the turning point in uncovering the eigenvalues of matrix \( A \), as solving \( \det(A - \lambda I)=0 \) results in the characteristic polynomial, which can then be tackled through methods such as the quadratic formula.
Quadratic Formula
The quadratic formula is a vital tool for solving quadratic equations, and it often comes into play when finding the eigenvalues of a matrix. Once a 2x2 matrix leads to a characteristic polynomial of the form \( \lambda^2 + b\lambda + c = 0 \), it is time to use the quadratic formula:
  • Quadratic Formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
The formula requires identifying the coefficients \( a \), \( b \), and \( c \) from the quadratic equation. In the context of our example, the quadratic polynomial \( \lambda^2 + 2\lambda - 9 = 0 \) dictates \( a = 1 \), \( b = 2 \), and \( c = -9 \). Plugging these into the quadratic formula assists in finding the eigenvalues, which in our case, turned out to be \( \lambda_1 = 1 + 3\sqrt{5} \) and \( \lambda_2 = 1 - 3\sqrt{5} \). These solutions are crucial, as they lead us to the corresponding eigenvectors associated with each eigenvalue.
Elementary Row Operations
Elementary row operations are basic transformations performed on matrices, which assist in simplifying them. These operations are particularly useful when solving systems of linear equations, including finding eigenvectors once eigenvalues are determined.There are three types of operations:
  • Swapping two rows
  • Multiplying a row by a non-zero number
  • Adding or subtracting a multiple of one row to another
When determining eigenvectors, row operations help manipulate the augmented matrix derived from \( (A - \lambda I)x = 0 \) to a simpler form. This process helps in identifying non-trivial solutions for the system. For instance, once \( A - \lambda I \) is established for each eigenvalue, the matrix is transformed using row operations to reveal the corresponding eigenvector. This step-by-step simplification is crucial in rendering the system solvable without altering the solution set, leading to insights about the directions dictated by the eigenvectors.

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Most popular questions from this chapter

In Theorem \(7.3 .3,\) we proved that similar matrices have the same eigenvalues. This problem investigates the relationship between their eigenvectors. Let \(\mathbf{v}\) be an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\) Prove that if \(B=S^{-1} A S,\) then \(S^{-1} \mathbf{v}\) is an eigenvector of \(B\) corresponding to the eigenvalue \(\lambda\).

Find the Jordan canonical form of each matrix. $$A=\left[\begin{array}{rrr} 2 & 1 & 1 \\ 2 & 1 & -2 \\ -1 & 0 & -2 \end{array}\right].$$ [Hint: The eigenvalues of \(A \text { are } \lambda=-1 \text { and } \lambda=3 .]\)

Let \(A\) be a nondefective matrix. Then $$S^{-1} A S=D,$$ where \(D\) is a diagonal matrix. This can be written as $$A=S D S^{-1},$$ Use this result to show that $$A^{2}=S D^{2} S^{-1},$$ and that for every positive integer \(k\) $$A^{k}=S D^{k} S^{-1}.$$

Use the result of Problem 32 to determine the sum and the product of the eigenvalues of the given matrix \(A\). $$A=\left[\begin{array}{rrrr} 0 & -3 & 1 & 1 \\ 0 & 2 & -1 & 3 \\ -1 & 1 & 1 & 1 \\ 1 & 0 & 5 & -2 \end{array}\right]$$

Use Jordan canonical forms to determine whether the given pair of matrices are similar. \(A=\left[\begin{array}{rrr}7 & -2 & 2 \\ 0 & 4 & -1 \\ -1 & 1 & 4\end{array}\right] ; B=\left[\begin{array}{rrr}3 & -1 & -2 \\ 1 & 6 & 1 \\\ 1 & 0 & 6\end{array}\right]\).
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