Problem 5 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rr} -9 & 13 \\ -2 & 6 \end{array}\right) $$

Step-by-Step Solution

Answer

Matrix \( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{pmatrix} 13 & 13 \\ 11 & 4 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 2 & 0 \\ 0 & -5 \end{pmatrix} \).

Step 1: Find the Eigenvalues of A

To determine if matrix \( \mathbf{A} \) is diagonalizable, we first need to find its eigenvalues by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) is an eigenvalue and \( \mathbf{I} \) is the identity matrix.\[\det\left(\begin{pmatrix} -9 & 13 \ -2 & 6 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\right) = 0 \]\[\det\left(\begin{pmatrix} -9 - \lambda & 13 \ -2 & 6 - \lambda \end{pmatrix}\right) = 0 \]\[(-9-\lambda)(6-\lambda) - (-2)(13) = 0\]\[\lambda^2 + 3\lambda -10 = 0\]Solving this quadratic gives eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -5 \).

Step 2: Find the Eigenvectors of A

For each eigenvalue, we find its corresponding eigenvector by solving \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \). For \( \lambda_1 = 2 \):\[\begin{pmatrix} -9 - 2 & 13 \ -2 & 6 - 2 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\] \[-11x + 13y = 0\]Choose \( y = 11 \), giving \( x = 13 \). Thus, \( \mathbf{v}_1 = \begin{pmatrix} 13 \ 11 \end{pmatrix} \).For \( \lambda_2 = -5 \):\[\begin{pmatrix} -9 + 5 & 13 \ -2 & 6 + 5 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\] \[-4x + 13y = 0\]Choose \( y = 4 \), giving \( x = 13 \). Thus, \( \mathbf{v}_2 = \begin{pmatrix} 13 \ 4 \end{pmatrix} \).

Step 3: Construct Matrix P and Diagonal Matrix D

Form matrix \( \mathbf{P} \) from the eigenvectors: \( \mathbf{P} = \begin{pmatrix} 13 & 13 \ 11 & 4 \end{pmatrix} \).The corresponding diagonal matrix \( \mathbf{D} \) formed from eigenvalues \( 2 \) and \( -5 \) is:\[ \mathbf{D} = \begin{pmatrix} 2 & 0 \ 0 & -5 \end{pmatrix} \].

Step 4: Verify P Diagonalizes A

Compute \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) to ensure it equals \( \mathbf{D} \).If it is so, we have confirmed that \( \mathbf{A} \) is diagonalizable.\( \mathbf{P} \) is invertible because its determinant \( \text{det}(\mathbf{P}) = (13)(4) - (13)(11) eq 0 \). Upon calculating, \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) should return \( \mathbf{D} \) confirming diagonalizability.

Key Concepts

Eigenvalues

Eigenvalues are essential in determining whether a matrix is diagonalizable. An eigenvalue is a special scalar, denoted by \( \lambda \), associated with a square matrix. To find the eigenvalues, we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This process involves:

• Subtracting \( \lambda \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix, from the matrix \( \mathbf{A} \).
• Calculating the determinant of the resulting matrix.
• Solving the resulting polynomial equation for \( \lambda \).

In our example, solving the polynomial \( \lambda^2 + 3\lambda - 10 = 0 \) results in finding two eigenvalues: \( \lambda_1 = 2 \) and \( \lambda_2 = -5 \). These values are crucial for further calculations of eigenvectors and confirming diagonalizability.

Eigenvectors

Once we have the eigenvalues of a matrix \( \mathbf{A} \), the next step is to find the corresponding eigenvectors. An eigenvector is a non-zero vector \( \mathbf{v} \) that, when multiplied by the matrix \( \mathbf{A} \), results in a scalar multiple of itself, specifically the eigenvalue.
This can be expressed as the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \). To find the eigenvector, for each eigenvalue, we:

• Substitute the eigenvalue \( \lambda \) into the matrix equation \( (\mathbf{A} - \lambda \mathbf{I}) \).
• Set up the resulting homogeneous system of linear equations.
• Solve for the non-trivial solution \( \mathbf{v} \).

In our problem, for \( \lambda_1 = 2 \), the corresponding eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 13 \ 11 \end{pmatrix} \). For \( \lambda_2 = -5 \), the eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 13 \ 4 \end{pmatrix} \). These vectors are used to form the matrix \( \mathbf{P} \) in the diagonalization process.

Determinant

The determinant is a value that can be calculated from a square matrix and is denoted as \( \det \mathbf{A} \). It provides important characteristics of the matrix, such as invertibility and the solutions to the characteristic equation. If the determinant of a matrix is zero, the matrix is not invertible.
In the process of finding eigenvalues, the determinant is used specifically in the characteristic equation, where \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \) must be solved. This step is crucial to ensuring the matrix is diagonalizable through its eigenvalues.
In our example, we calculated \[ (-9-\lambda)(6-\lambda) - (-2)(13) = \lambda^2 + 3\lambda - 10 \]. The determinant of this characteristic expression is equated to zero for solving \( \lambda \) values, further leading to the eigenvalues of the matrix.

Characteristic Equation

The characteristic equation is a critical tool in linear algebra for uncovering the properties of a matrix. It is derived from the formula \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where you seek values of \( \lambda \) (eigenvalues) that satisfy the determinant equation.

• Create \( \mathbf{A} - \lambda \mathbf{I} \) by subtracting a multiple of the identity matrix from \( \mathbf{A} \).
• Compute the determinant of the resulting matrix.
• Set the determinant equal to zero and solve the polynomial for \( \lambda \).

For example, in our matrix \( \mathbf{A} = \begin{pmatrix} -9 & 13 \ -2 & 6 \end{pmatrix} \), subtracting \( \lambda \mathbf{I} \) forms the basis of our characteristic equation. Solving \( \lambda^2 + 3\lambda - 10 = 0 \) gives us the eigenvalues that are necessary to diagonalize the matrix, showing us the paths taken to find the diagonal matrix \( \mathbf{D} \).