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

Step-by-Step Solution

Answer

The matrix \(\mathbf{A}\) is not diagonalizable.

Step 1: Find the Eigenvalues

First, find the eigenvalues of the matrix \(\mathbf{A}\) by solving the characteristic polynomial, which is obtained from the determinant of \(\mathbf{A} - \lambda \mathbf{I}\). Compute \[\text{det}(\mathbf{A} - \lambda \mathbf{I}) = \begin{vmatrix} -5-\lambda & -3 \ 5 & 11-\lambda \end{vmatrix}=0.\] By solving \((-5-\lambda)(11-\lambda) + 15 = 0\), you get the eigenvalues \(\lambda_1 = 3\) and \(\lambda_2 = 3\).

Step 2: Check Algebraic and Geometric Multiplicity

Since both eigenvalues are the same, we have one repeated eigenvalue. For \(\mathbf{A}\) to be diagonalizable, the geometric multiplicity must equal the algebraic multiplicity. Here, the algebraic multiplicity is 2. Next, find the geometric multiplicity by solving \((\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0}\) to find the eigenvector space for \(\lambda = 3\).

Step 3: Solve for Eigenvectors

The matrix \(\mathbf{A} - 3\mathbf{I}\) becomes \(\begin{pmatrix} -8 & -3 \ 5 & 8 \end{pmatrix}\). Finding the null space (solutions to the system) gives \(\mathbf{v} = k \begin{pmatrix} 1 \ -\frac{8}{5} \end{pmatrix}\) for some scalar \(k\). This means the geometric multiplicity is 1.

Step 4: Determine if Diagonalizable

Since the geometric multiplicity is 1 but the algebraic multiplicity is 2, \(\mathbf{A}\) cannot have a complete set of linearly independent eigenvectors, and thus it is not diagonalizable.

Key Concepts

Eigenvalues

When working with matrices, eigenvalues are one of the key concepts used to understand the structure of a matrix. To find the eigenvalues, we solve the characteristic equation, which originates from the determinant of the matrix \(\mathbf{A} - \lambda \mathbf{I}\), where \(\lambda\) represents an eigenvalue and \(\mathbf{I}\) is the identity matrix.
This determinant becomes a polynomial equation, known as the characteristic polynomial, and solving it yields the eigenvalues. For example, for the matrix given in our problem:

\[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = \begin{vmatrix} -5-\lambda & -3 \ 5 & 11-\lambda \end{vmatrix}=0. \]
By calculating this, we found that both eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = 3\).

• Eigenvalues are crucial because they describe the "scale" of the matrix’s transformation effect.
• Their repeated nature in this problem leads to more investigation of their multiplicities.

Geometric Multiplicity

Geometric multiplicity is another important property of eigenvalues that describes the number of linearly independent eigenvectors associated with each eigenvalue. It is computed by finding the null space of \(\mathbf{A} - \lambda \mathbf{I}\).
For the eigenvalue \(\lambda = 3\), we form the matrix:

\[\mathbf{A} - 3\mathbf{I} = \begin{pmatrix} -8 & -3 \ 5 & 8 \end{pmatrix}\]
To find the null space, we solve the equation:
\[(\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0}\]
Upon solving, we find that there is only one linearly independent eigenvector:
\[\mathbf{v} = k \begin{pmatrix} 1 \ -\frac{8}{5} \end{pmatrix}\]

• The geometric multiplicity here is 1, indicating that there is only one independent direction the transformation described by the matrix scales by a factor of 3.

Algebraic Multiplicity

Algebraic multiplicity refers to how many times an eigenvalue appears as a root of the characteristic polynomial. It tells us the frequency of each eigenvalue but does not indicate the number of linearly independent eigenvectors.
In this example, the characteristic polynomial factors to yield the eigenvalue \(\lambda = 3\) twice, so its algebraic multiplicity is 2.

• Having a higher algebraic multiplicity suggests the eigenvalue should have richer eigenvector behavior, contingent on matching geometric multiplicity.

This distinction is critical when considering diagonalizability. A matrix is diagonalizable if, for every eigenvalue, the geometric and algebraic multiplicities match. Since our eigenvalue \(\lambda = 3\) has a geometric multiplicity of 1, but an algebraic multiplicity of 2, the matrix \(\mathbf{A}\) cannot be diagonalized as we don't have enough directions (eigenvectors) to span the space.