Problem 7 - Advanced Engineering Mathematics

Question

In Problems 1-20, 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{ll} \frac{1}{2} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{2} \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix is diagonalizable with eigenvalues \( \frac{2}{3} \) and \( \frac{1}{3} \).

Step 1: Find the Eigenvalues

Determine the eigenvalues by solving the characteristic equation: \\[ \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \] where \( \mathbf{I} \) is the identity matrix. For the given matrix \( \mathbf{A} \), this equation becomes: \\[ \left( \frac{1}{2} - \lambda \right) \left( \frac{1}{2} - \lambda \right) - \left( \frac{1}{6} \right)^2 = 0 \] \[ \left( \frac{1}{2} - \lambda \right)^2 - \frac{1}{36} = 0 \] \[ \left( \frac{1}{4} - \lambda + \lambda^2 \right) - \frac{1}{36} = 0 \] Simplify to find: \[ \lambda^2 - \lambda + \frac{1}{4} - \frac{1}{36} = 0 \] Solve this quadratic equation to get the eigenvalues \( \lambda_1 \) and \( \lambda_2 \).

Step 2: Solve the Quadratic Equation

Simplify the equation from Step 1: \\[ \lambda^2 - \lambda + \frac{8}{36} - \frac{1}{36} = 0 \] which simplifies to: \\[ \lambda^2 - \lambda + \frac{7}{36} = 0 \] Use the quadratic formula: \\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -1, c = \frac{7}{36} \). Calculate to find \( \lambda_1 = \frac{2}{3} \) and \( \lambda_2 = \frac{1}{3} \).

Step 3: Find Eigenvectors for Each Eigenvalue

For each eigenvalue found in Step 2, solve \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0 \) to find the corresponding eigenvector \( \mathbf{v} \). \[ (\mathbf{A} - \lambda_1 \mathbf{I}) = \begin{pmatrix} \frac{1}{2} - \frac{2}{3} & \frac{1}{6} \ \frac{1}{6} & \frac{1}{2} - \frac{2}{3} \end{pmatrix} \] Simplify to get eigenvectors for \( \lambda_1 = \frac{2}{3} \) and \( \lambda_2 = \frac{1}{3} \).

Step 4: Form the Matrix P

Form the matrix \( \mathbf{P} \) using the eigenvectors from Step 3 as its columns. Ensure that the order of the eigenvectors matches the order of the eigenvalues in the resulting diagonal matrix \( \mathbf{D} \).

Step 5: Form the Diagonal Matrix D

Create a diagonal matrix \( \mathbf{D} \) where the diagonal elements are the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) found in Step 2, placed in the same order corresponding to \( \mathbf{P} \).

Step 6: Verify the Diagonalization

Check if \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) holds. Compute \( \mathbf{P}^{-1} \) and then calculate \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \). Ensure the result is equivalent to \( \mathbf{D} \).

Key Concepts

Eigenvalues

Eigenvalues are numerical values that provide essential information about a matrix's properties. They are a measure of a matrix's ability to stretch or compress vectors. To find eigenvalues, we must solve the characteristic equation of the matrix.
This is achieved by evaluating the determinant of the matrix subtracted by the identity matrix multiplied by a scalar \( \lambda \), which leads to \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \).
In simple terms, solving for \( \lambda \) allows us to uncover points where transformation vectors don't change direction, which are the eigenvalues. For our matrix, the eigenvalues turned out to be \( \frac{2}{3} \) and \( \frac{1}{3} \).

Eigenvectors

Eigenvectors are vectors associated with eigenvalues that, when a matrix operates on them, change only in scale, not in direction. They are crucial because they reveal invariant subspaces under matrix transformations, giving insight into the matrix's intrinsic geometric properties.
To find an eigenvector for a given eigenvalue, we solve the equation \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0\) for the vector \(\mathbf{v}\).
This process entails forming a system of linear equations which, when solved, gives us the specific direction (or directions) of invariance for each eigenvalue. For our specific case, this step provides the vectors necessary for diagonalization.

Characteristic Equation

The characteristic equation is a fundamental polynomial equation derived from a matrix, pivotal to finding eigenvalues. It appears when setting the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) equal to zero. This equation encodes the matrix's property transformations and allows us to find \( \lambda \), the eigenvalues.
For example, with our matrix, the equation \( \left( \frac{1}{2} - \lambda \right)^2 - \left( \frac{1}{6} \right)^2 = 0 \) is the specific characteristic equation used to find our eigenvalues.
Solving this polynomial equation helps identify eigenvalues, crucial for understanding the matrix's diagonalization potential.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). Commonly used to find eigenvalues from characteristic equations, the formula is: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a \), \( b \), and \( c \) represent coefficients from the characteristic equation. In our matrix's example, we identified \( a = 1 \), \( b = -1 \), and \( c = \frac{7}{36} \). Plugging these values into the quadratic formula, yielded the solutions \( \lambda_1 = \frac{2}{3} \) and \( \lambda_2 = \frac{1}{3} \).
This method is reliable for efficiently locating the roots of a quadratic equation, and thus finding the eigenvalues necessary for matrix diagonalization.