Problem 1 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right) $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 1 \end{pmatrix} \).

Step 1: Calculate the Characteristic Polynomial

To determine if the matrix \( \mathbf{A} \) is diagonalizable, we first find its characteristic polynomial. The characteristic polynomial is given by \( \det(\mathbf{A} - \lambda \mathbf{I}) \), where \( \lambda \) represents the eigenvalues and \( \mathbf{I} \) is the identity matrix. Set up the equation: \[det\left(\begin{array}{cc} 2-\lambda & 3 \ 1 & 4-\lambda \end{array}\right) = 0\]Calculate the determinant: \[(2-\lambda)(4-\lambda) - 3 \times 1 = \lambda^2 - 6\lambda + 5\]

Step 2: Solve the Characteristic Equation

Solve \( \lambda^2 - 6\lambda + 5 = 0 \) to find the eigenvalues. This is a quadratic equation that can be solved using the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our equation, \( a = 1 \), \( b = -6 \), \( c = 5 \). Plug these into the formula:\[\lambda = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 5}}{2}\]\[\lambda = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2}\]\[\lambda = \frac{6 \pm 4}{2}\]This gives two eigenvalues: \( \lambda_1 = 5 \) and \( \lambda_2 = 1 \).

Step 3: Find the Eigenvectors

For each eigenvalue, find the corresponding eigenvector. Start with \( \lambda_1 = 5 \):Substitute into \( \mathbf{A} - \lambda \mathbf{I} \), giving:\[\begin{pmatrix} 2-5 & 3 \ 1 & 4-5 \end{pmatrix} = \begin{pmatrix} -3 & 3 \ 1 & -1 \end{pmatrix}\]Find a vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) such that \( (\mathbf{A} - \lambda_1 \mathbf{I}) \mathbf{v} = \mathbf{0} \):\[-3x + 3y = 0\]\[x = y\]One eigenvector is \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \).Repeat for \( \lambda_2 = 1 \):\[\begin{pmatrix} 2-1 & 3 \ 1 & 4-1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \ 1 & 3 \end{pmatrix}\]Solve:\[1x + 3y = 0 \Rightarrow x = -3y\]An eigenvector is \( \begin{pmatrix} -3 \ 1 \end{pmatrix} \).

Step 4: Form the Matrix \( \mathbf{P} \) and Diagonal Matrix \( \mathbf{D} \)

The matrix \( \mathbf{P} \) is formed by the eigenvectors \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \) and \( \begin{pmatrix} -3 \ 1 \end{pmatrix} \) as columns:\[\mathbf{P} = \begin{pmatrix} 1 & -3 \ 1 & 1 \end{pmatrix}\]The diagonal matrix \( \mathbf{D} \) will have the eigenvalues on the diagonal:\[\mathbf{D} = \begin{pmatrix} 5 & 0 \ 0 & 1 \end{pmatrix}\]

Step 5: Verify the Diagonalization

Verify if \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \). First, find \( \mathbf{P}^{-1} \).For \( \mathbf{P} = \begin{pmatrix} 1 & -3 \ 1 & 1 \end{pmatrix} \), calculate the inverse:The determinant of \( \mathbf{P} \) is \( (1)(1) - (-3)(1) = 4 \).So, \( \mathbf{P}^{-1} = \frac{1}{4} \begin{pmatrix} 1 & 3 \ -1 & 1 \end{pmatrix} \).Check the multiplication \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) to ensure it equals \( \mathbf{D} \). If satisfied, the matrix is diagonalizable.

Key Concepts

Eigenvalues

In the context of matrices, eigenvalues are crucial for understanding how a matrix can be transformed through diagonalization. An eigenvalue, represented by \( \lambda \), is a special number associated with a matrix \( \mathbf{A} \). It comes about through the equation \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \), where \( \mathbf{v} \) is a non-zero vector known as an eigenvector. Perhaps the simplest way to think about an eigenvalue is as a "scalar" that stretches or compresses an eigenvector during the transformation induced by \( \mathbf{A} \).
To compute eigenvalues, we use the characteristic polynomial, which we'll learn about soon. Solving for \( \lambda \) means finding the roots of this polynomial. Once these roots are identified, diagonalization of the matrix is possible if the matrix meets certain mathematical criteria, specifically, having distinct eigenvalues.

Eigenvectors

Eigenvectors are non-zero vectors that only change in scale when a linear transformation is applied to them via a matrix. These vectors play a key role in matrix diagonalization. Given a matrix \( \mathbf{A} \) and an eigenvalue \( \lambda \), the corresponding eigenvector \( \mathbf{v} \) can be found from the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \). This equation underscores a fundamental property: eigenvectors correspond to particular eigenvalues.
Finding the eigenvectors involves solving a system of linear equations derived when substituting the eigenvalues into the expression \( \mathbf{A} - \lambda \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix. The solutions to this system give the direction of the vector \( \mathbf{v} \), reiterating its nature as a vector that remains stable under transformation except for scaling.

Characteristic Polynomial

The characteristic polynomial of a matrix is key to finding the eigenvalues. It is formulated from the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This polynomial has degree equal to the dimension of the square matrix \( \mathbf{A} \). Its roots are precisely the eigenvalues of the matrix.
In our step-by-step solution, we began with the matrix \( \mathbf{A} \) and substituted into this equation, resulting in a quadratic polynomial \( \lambda^2 - 6\lambda + 5 \). Solving this polynomial involved utilizing the quadratic formula, leading to the eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = 1 \). Identifying these eigenvalues is the first crucial step in determining whether a matrix is diagonalizable.

Diagonal Matrix

A diagonal matrix, often represented by \( \mathbf{D} \), where all off-diagonal elements are zero, simplifies matrix operations significantly. Diagonalization is a process through which we express a matrix \( \mathbf{A} \) as \( \mathbf{PDP}^{-1} \), with \( \mathbf{D} \) being the diagonal matrix that includes the eigenvalues along its main diagonal.
This structural form is advantageous because diagonal matrices are easier to raise to powers and compute functions such as exponentials. In the step-by-step problem, once we discovered the eigenvalues and corresponding eigenvectors, we constructed the matrix \( \mathbf{P} \) with eigenvectors as columns. With these in hand, \( \mathbf{D} \) can be formed, showcasing its simplicity and pivotal role in matrix algebra.