Problem 8 - Advanced Engineering Mathematics

Question

In Problems, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rr} -2 & -1 \\ 1 & -4 \end{array}\right) $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{A} \) is diagonalizable. \( \mathbf{P} = \begin{pmatrix} 1 - \sqrt{2} & 1 + \sqrt{2} \\ 1 & 1 \end{pmatrix} \), \( \mathbf{D} = \begin{pmatrix} -3+\sqrt{2} & 0 \\ 0 & -3-\sqrt{2} \end{pmatrix} \).

Step 1: Calculate the Characteristic Polynomial

To check if a matrix is diagonalizable, we start by computing its characteristic polynomial. The characteristic polynomial of a matrix \( \mathbf{A} \) is given by \( \det(\mathbf{A} - \lambda \mathbf{I}) \), where \( \mathbf{I} \) is the identity matrix.\[\mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} -2 - \lambda & -1 \ 1 & -4 - \lambda \end{pmatrix}\]The determinant is:\[\det(\mathbf{A} - \lambda \mathbf{I}) = (-2 - \lambda)(-4 - \lambda) - (-1)(1)\]Simplifying:\[= \lambda^2 + 6\lambda + 7\]

Step 2: Find Eigenvalues

Eigenvalues are obtained by solving the characteristic equation \( \lambda^2 + 6\lambda + 7 = 0 \). Use the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our polynomial, \( a = 1 \), \( b = 6 \), \( c = 7 \), thus:\[\lambda = \frac{-6 \pm \sqrt{36 - 28}}{2} = \frac{-6 \pm \sqrt{8}}{2}\]\[= \frac{-6 \pm 2\sqrt{2}}{2} = -3 \pm \sqrt{2}\]The eigenvalues are \( \lambda_1 = -3 + \sqrt{2} \) and \( \lambda_2 = -3 - \sqrt{2} \).

Step 3: Verify Diagonalizability

A matrix is diagonalizable if it has \( n \) linearly independent eigenvectors for an \( n \times n \) matrix. Here, \( \mathbf{A} \) is \( 2 \times 2 \) and has distinct eigenvalues \( \lambda_1 \) and \( \lambda_2 \). Therefore, \( \mathbf{A} \) is diagonalizable.

Step 4: Find Eigenvectors

To find the eigenvectors, consider the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0} \). Solve this for both eigenvalues.**For \( \lambda_1 = -3 + \sqrt{2} \):**\[\mathbf{A} - \lambda_1 \mathbf{I} = \begin{pmatrix} -2 - (-3 + \sqrt{2}) & -1 \ 1 & -4 - (-3 + \sqrt{2}) \end{pmatrix} = \begin{pmatrix} 1 - \sqrt{2} & -1 \ 1 & -1 - \sqrt{2} \end{pmatrix}\]The system of equations provided by this matrix is:\[(1 - \sqrt{2})x - y = 0\]\[x - (1 + \sqrt{2})y = 0\]Choose \( y = 1 \), then \( x = 1 - \sqrt{2} \ \Rightarrow \mathbf{v_1} = \begin{pmatrix} 1 - \sqrt{2} \ 1 \end{pmatrix} \).**For \( \lambda_2 = -3 - \sqrt{2} \):**\[\mathbf{A} - \lambda_2 \mathbf{I} = \begin{pmatrix} -2 - (-3 - \sqrt{2}) & -1 \ 1 & -4 - (-3 - \sqrt{2}) \end{pmatrix} = \begin{pmatrix} 1 + \sqrt{2} & -1 \ 1 & -1 + \sqrt{2} \end{pmatrix}\]The system is:\[(1 + \sqrt{2})x - y = 0\]\[x - (1 - \sqrt{2})y = 0\]Choose \( y = 1 \), then \( x = 1 + \sqrt{2} \ \Rightarrow \mathbf{v_2} = \begin{pmatrix} 1 + \sqrt{2} \ 1 \end{pmatrix} \).

Step 5: Form Matrix P and Diagonal Matrix D

The matrix \( \mathbf{P} \) is formed from the eigenvectors \( \mathbf{v_1} \) and \( \mathbf{v_2} \):\[\mathbf{P} = \begin{pmatrix} 1 - \sqrt{2} & 1 + \sqrt{2} \ 1 & 1 \end{pmatrix}\]The diagonal matrix \( \mathbf{D} \) contains the eigenvalues on its diagonal:\[\mathbf{D} = \begin{pmatrix} -3 + \sqrt{2} & 0 \ 0 & -3 - \sqrt{2} \end{pmatrix}\]

Step 6: Verify the Diagonalization

To verify, ensure that \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \). Calculate \( \mathbf{P}^{-1} \) and confirm:Calculate \( \mathbf{P}^{-1} \):The inverse of a 2x2 matrix \( \mathbf{P} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is:\[\mathbf{P}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]For \( \mathbf{P} = \begin{pmatrix} 1 - \sqrt{2} & 1 + \sqrt{2} \ 1 & 1 \end{pmatrix} \):\[\mathbf{P}^{-1} = \frac{1}{2\sqrt{2}} \begin{pmatrix} 1 & -1 - \sqrt{2} \ -1 & 1 - \sqrt{2} \end{pmatrix}\]Verify by calculating \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) to see if it equals \( \mathbf{D} \). This computation will confirm if \( \mathbf{A} \) is diagonalizable with \( \mathbf{D} \) as its diagonalized form.

Key Concepts

Eigenvalues

Eigenvalues are special numbers that originate from a square matrix. By determining these values, we can understand the essence of the transformations a matrix represents.
An eigenvalue, often denoted as \( \lambda \), is derived from the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = \mathbf{0} \), where \( \mathbf{A} \) is the matrix in question, \( \lambda \) is the eigenvalue, and \( \mathbf{I} \) is the identity matrix.

• Eigenvalues tell us how a transformation stretches or compresses along various directions.
• The magnitude of an eigenvalue indicates the scale of transformation.

In our example, the characteristic polynomial revealed the eigenvalues \( \lambda_1 = -3 + \sqrt{2} \) and \( \lambda_2 = -3 - \sqrt{2} \). These distinct values suggest that the matrix can be diagonalized. Understanding eigenvalues is crucial as they provide insights into matrix behaviors in simplified, diagonal forms.

Eigenvectors

Eigenvectors are vectors that remain in the same span after a matrix transformation, merely scaled by the corresponding eigenvalue.
For a given eigenvalue \( \lambda \), an eigenvector \( \mathbf{v} \) satisfies the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \). Here, \( \mathbf{A} \) is our matrix and \( \mathbf{I} \) is the identity matrix.

• Eigenvectors represent directions preserved during transformations.
• Determining eigenvectors involves solving a system of linear equations.

In this problem, we found the eigenvectors by calculating for each eigenvalue. For \( \lambda_1 = -3 + \sqrt{2} \), the eigenvector is \( \mathbf{v_1} = \begin{pmatrix} 1 - \sqrt{2} \ 1 \end{pmatrix} \). For \( \lambda_2 = -3 - \sqrt{2} \), it is \( \mathbf{v_2} = \begin{pmatrix} 1 + \sqrt{2} \ 1 \end{pmatrix} \). The presence of two linearly independent eigenvectors reaffirms the diagonalizability of the matrix \( \mathbf{A} \).

Characteristic Polynomial

The characteristic polynomial is a function that encodes information about a matrix's eigenvalues, providing a powerful tool for understanding the matrix's properties.
To find it, we compute \( \det(\mathbf{A} - \lambda \mathbf{I}) \), which leads us to a polynomial in terms of \( \lambda \).

• This polynomial is crucial for identifying eigenvalues.
• Its degree matches the dimension of the matrix.

In our example, the characteristic polynomial of the matrix \( \mathbf{A} \) was found as \( \lambda^2 + 6\lambda + 7 \). Solving this polynomial equation using the quadratic formula yielded the eigenvalues. The solutions to the polynomial determine not only whether a matrix can be diagonalized but also the nature of its eigenvectors and eigenvalues.

Quadratic Formula

The quadratic formula is an essential mathematical tool for solving quadratic equations, frequently appearing when finding eigenvalues.
The formula is expressed as \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).

• It provides a straightforward method to solve any quadratic equation.
• The discriminant \( b^2 - 4ac \) determines the nature and number of solutions.

For our matrix example, the characteristic polynomial \( \lambda^2 + 6\lambda + 7 \) was solved using this formula. With coefficients \( a = 1 \), \( b = 6 \), and \( c = 7 \), we calculated \( \lambda = \frac{-6 \pm \sqrt{8}}{2} \), leading to the eigenvalues. Utilizing the quadratic formula simplifies computations and is invaluable for determining matrix properties.