Question

Given that the matrix $$ \mathrm{A}=\left(\begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right) $$ has two eigenvectors of the form \((1 \quad y \quad 1)^{\mathrm{T}}\), use the stationary property of the expression \(J(\mathrm{x})=\mathrm{x}^{\mathrm{T}} \mathrm{Ax} /\left(\mathrm{x}^{\mathrm{T}} \mathrm{x}\right)\) to obtain the corresponding eigenvalues. Deduce the third eigenvalue.

Short answer

The corresponding eigenvalues are \( \lambda_1 = \lambda_2 = \frac{2}{3} \) and the third eigenvalue is \( \lambda_3 = \frac{14}{3} \).

Step-by-step solution

Set Up the Stationary Property

The function given is \(J(\textbf{x}) = \frac{\textbf{x}^{\text{T}} \textbf{Ax}}{\textbf{x}^{\text{T}} \textbf{x}}\). To find the eigenvalues using the stationary property, set \( abla J(\textbf{x}) = 0 \)

Compute \(J(\textbf{x})\) for the Eigenvector

Given the eigenvector form \( \textbf{x} = (1, y, 1)^{T} \), substitute \( \textbf{x} \) into \( J(\textbf{x}) \). First, calculate \( \textbf{Ax} \): \ \[ \textbf{A} (1, y, 1)^{\text{T}} = \left(\begin{array}{ccc} 2 & -1 & 0 \ -1 & 2 & -1 \ 0 & -1 & 2 \end{array}\right) \left(\begin{array}{c} 1 \ y \ 1 \end{array}\right) = \left(\begin{array}{c} 2 - y \ -1 + 2y - 1 \ -y + 2 \end{array}\right) = \left(\begin{array}{c} 2 - y \ 2y - 2 \ 2 - y \end{array}\right) \]

Simplify the Expression

Simplify the quadratic form \( \textbf{x}^{\text{T}} \textbf{Ax} \): \ \[ \textbf{x}^{\text{T}} \textbf{Ax} = (1, y, 1) \left(\begin{array}{c} 2 - y \ 2y - 2 \ 2 - y \end{array}\right) = 1(2 - y) + y(2y - 2) + 1(2 - y) \] \ \[= (2 - y) + (2y^2 - 2y) + (2 - y) = 2 - y + 2y^2 - 2y + 2 - y = 2y^2 - 4y + 4\]

Set Up the Denominator

Compute \( \textbf{x}^{\text{T}} \textbf{x} \) : \ \[ (1, y, 1)^{\text{T}} (1, y, 1) = 1^2 + y^2 + 1^2 = y^2 + 2 \]

Form the Rayleigh Quotient

Plug \( \textbf{x}^{\text{T}} \textbf{Ax} \) and \( \textbf{x}^{\text{T}} \textbf{x} \) into \( J(\textbf{x}) \): \ \[ J(\textbf{x}) = \frac{2y^2 - 4y + 4}{y^2 + 2} \]

Find the Derivative

To find the stationary points, set the derivative of \( J(\textbf{x}) \) with respect to \( y \) to zero: \ \[ \frac{d}{dy} \left( \frac{2y^2 - 4y + 4}{y^2 + 2} \right) = 0 \] \ Use the quotient rule: \ \[ \left( \frac{(y^2 + 2)(4y - 4) - (2y^2 - 4y + 4)(2y)}{(y^2 + 2)^2} \right) = 0 \] \ \[ \rightarrow (y^2 + 2)(4y - 4) - 2y(2y^2 - 4y + 4) = 0 \] \ \[ \rightarrow 4y^3 - 6y^2 - 4y + 8 = 0 \]

Solve the Polynomial

Solve for \( y \): This cubic equation simplifies to \( (y-1)^2(2y-4) = 0 \), so the solutions are \( y = 1 \) and \( y = 2 \). These correspond to two eigenvalues.

Calculate Corresponding Eigenvalues

For \( y = 1 \): \ \[ J(\textbf{x}) = \frac{2(1)^2 - 4(1) + 4}{(1)^2 + 2} = \frac{2 - 4 + 4}{1 + 2} = \frac{2}{3} \rightarrow \lambda_1 = \frac{2}{3}\] \ For \( y = 2 \): \ \[ J(\textbf{x}) = \frac{2(2)^2 - 4(2) + 4}{(2)^2 + 2} = \frac{8 - 8 + 4}{4 + 2} = \frac{4}{6} = \frac{2}{3} \rightarrow \lambda_2 = \frac{2}{3} \]

Deduce the Third Eigenvalue

Given that the trace of matrix \( \textbf{A} \) (i.e., sum of its diagonal elements) is equal to the sum of its eigenvalues, calculate the third eigenvalue: \ The trace is: \( 2 + 2 + 2 = 6 \). \ Therefore, the third eigenvalue: \ \[ \lambda_1 + \lambda_2 + \lambda_3 = 6 \rightarrow \frac{2}{3} + \frac{2}{3} + \lambda_3 = 6 \] \ \[ \lambda_3 = 6 - \frac{4}{3} = \frac{14}{3} \]

Key concepts

matrix eigenvalues

Matrix eigenvalues are fundamental concepts in linear algebra. They can tell us about the essential properties of a matrix. An eigenvalue of a matrix is a scalar that shows the factor by which an eigenvector is scaled when the matrix is applied to it. To find eigenvalues, we typically solve the characteristic equation: \(\mathbf{A} \mathbf{v} = \lambda \mathbf{v}\), where \(\mathbf{A}\) is our matrix, \(\mathbf{v}\) is the eigenvector, and \(\lambda\) is the eigenvalue. Eigenvalues help us understand stability in systems, vibrations in mechanical structures, and many other applications in physics and engineering.

Rayleigh quotient

The Rayleigh quotient is a practical tool for computing eigenvalues. Specifically, for a vector \(\mathbf{x}\) and a matrix \(\mathbf{A}\), it is defined as: \(J(\mathbf{x}) = \frac{\mathbf{x}^{\mathrm{T}} \mathbf{Ax}}{\mathbf{x}^{\mathrm{T}} \mathbf{x}}\). This function can be used to find eigenvalues when \(J(\mathbf{x})\) is at its stationary points, meaning its derivative equals zero. In our problem, the given eigenvector form \(\mathbf{x} = (1, y, 1)^{\mathrm{T}}\) is used to simplify and solve for the stationary points, thus giving us the eigenvalues. The Rayleigh quotient is valuable for finding the dominant eigenvalue and helps in various computational methods for eigenvalues.

eigenvectors

Eigenvectors are non-zero vectors that change at most by a scalar factor when a linear transformation is applied. Formally, for a matrix \(\mathbf{A}\) and eigenvalue \(\lambda\), the eigenvector \(\mathbf{v}\) satisfies the equation: \(\mathbf{A} \mathbf{v} = \lambda \mathbf{v}\). In the given exercise, the matrix \(\mathbf{A}\) has eigenvectors of the form \(\mathbf{x} = (1, y, 1)^{\mathrm{T}}\). These eigenvectors allow us to use the Rayleigh quotient to find the matrix's eigenvalues. Each eigenvector pairs with a corresponding eigenvalue, indicating the direction and scaling factor of the transformation described by the matrix. Understanding eigenvectors is crucial in many fields, from quantum mechanics to facial recognition technology.