Problem 6 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rr} -5 & -3 \\ 5 & 11 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix \(\mathbf{A}\) is not diagonalizable over the real numbers.

Step 1: Find the Eigenvalues of Matrix A

To determine if matrix \(\mathbf{A}\) is diagonalizable, we first need to find its eigenvalues. The eigenvalues \(\lambda\) are found by solving the characteristic equation \(\text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0\). For \(\mathbf{A} = \begin{pmatrix} -5 & -3 \ 5 & 11 \end{pmatrix}\), compute \(\text{det}(\begin{pmatrix} -5 - \lambda & -3 \ 5 & 11 - \lambda \end{pmatrix}) = 0\). This results in the equation \((-5 - \lambda)(11 - \lambda) + 15 = 0\). Simplifying, \(\lambda^2 - 6\lambda + 40 = 0\).

Step 2: Solve the Characteristic Equation

Now, we solve the quadratic equation \(\lambda^2 - 6\lambda + 40 = 0\) using the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 40\). Substituting these into the formula: \(\lambda = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 40}}{2}\). Simplifying the discriminant, \(\lambda = \frac{6 \pm \sqrt{36 - 160}}{2}\), we find \(\lambda = \frac{6 \pm \sqrt{-124}}{2}\), giving complex roots. So, the eigenvalues are complex.

Step 3: Determine Diagonalizability

A matrix is diagonalizable if it has \(n\) linearly independent eigenvectors for an \(n \times n\) matrix, where the eigenvectors correspond to distinct eigenvalues. Since the eigenvalues of \(\mathbf{A}\) are complex (and not distinct real numbers), $$\mathbf{A}$$ is not diagonalizable over the field of real numbers.

Key Concepts

Eigenvalues

To determine whether a matrix is diagonalizable, it starts with finding the eigenvalues. Eigenvalues, denoted usually by \( \lambda \), are special numbers that arise from matrices. They are obtained by solving the characteristic equation of the matrix, which is derived from \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For the given matrix \( \mathbf{A} = \begin{pmatrix} -5 & -3 \ 5 & 11 \end{pmatrix} \), we solve this by setting up the matrix \( \mathbf{A} - \lambda \mathbf{I} \) and calculating its determinant. For this exercise, the equation formed is \( (\lambda^2 - 6\lambda + 40 = 0) \).

The discriminant of the quadratic equation \( b^2 - 4ac \) is crucial. If it's negative, as in this case, the roots (eigenvalues) will be complex numbers. Complex eigenvalues suggest that diagonalization might not be possible over the real numbers. We'll need to check more details in the context of specific problems.

Eigenvectors

After finding eigenvalues, eigenvectors are calculated. An eigenvector corresponds to a matrix's eigenvalue and provides direction that doesn't change when the matrix transforms it. Given an eigenvalue \( \lambda \), we find the eigenvectors \( \mathbf{v} \) by solving \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = 0 \).

In simple terms, once you have an eigenvalue, plug it into the equation \( \mathbf{A} - \lambda \mathbf{I} \) and solve for the eigenvector. These vectors need to be linearly independent for diagonalization. With complex eigenvalues, as in the problem for matrix \( \mathbf{A} \), the eigenvectors could also be complex, influencing diagonalization possibilities.

Characteristic Equation

The characteristic equation is a polynomial equation derived from setting the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) equal to zero. It plays a central role in finding eigenvalues of a matrix.

• For a 2x2 matrix \( \mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the characteristic equation is \( \lambda^2 - \text{tr}(\mathbf{A})\lambda + \text{det}(\mathbf{A}) = 0 \).
• The trace \( \text{tr}(\mathbf{A}) \) is \( a + d \), and the determinant \( \text{det}(\mathbf{A}) \) is \( ad - bc \).

In our exercise, this resulted in \( \lambda^2 - 6\lambda + 40 = 0 \), a quadratic equation with complex solutions. The nature of these solutions often suggests whether or not diagonalization over real numbers is feasible.

Non-diagonalizable Matrix

A matrix is non-diagonalizable if it cannot be transformed into a diagonal matrix using a similarity transformation. This usually happens when a matrix does not have enough linearly independent eigenvectors.

For the exercise in question, matrix \( \mathbf{A} \) has complex eigenvalues. For a matrix to be diagonalizable over real numbers, its eigenvalues should not only be real but also distinct to ensure sufficient number of linearly independent eigenvectors.

• If eigenvalues are complex, as they are here, it limits possibilities for certain kinds of diagonalization (specifically over the reals).
• This results in \( \mathbf{A} \) being termed non-diagonalizable in the given field (the real numbers here).

Understanding this concept helps us see why some matrices refuse to be simplified fully into a diagonal form.