Problem 14 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{rrr} 0 & -9 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Determine Eigenvalues

First, find the eigenvalues of the matrix \(\mathbf{A}\). We need to solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{I}\) is the identity matrix. Calculate \(\mathbf{A} - \lambda \mathbf{I}\):\[\begin{pmatrix}-\lambda & -9 & 0 \1 & -\lambda & 0 \0 & 0 & 1-\lambda \\end{pmatrix}\]Compute the determinant:\[(-\lambda)((-\lambda)(1-\lambda) - 0 \cdot 0) - (-9)(0)\]This simplifies to \(-\lambda(-\lambda^2 + \lambda)\). Set the determinant to zero:\[-\lambda(\lambda^2 - \lambda) = 0\]The solutions are \(\lambda = 0, 0, 1\), giving us eigenvalues \(\lambda_1 = 0\) with multiplicity 2, and \(\lambda_2 = 1\).

Step 2: Find Eigenvectors for Each Eigenvalue

For \(\lambda_1 = 0\), solve \((\mathbf{A} - 0 \mathbf{I})\mathbf{x} = \mathbf{0}\), simplifying to:\[\begin{pmatrix}0 & -9 & 0 \1 & 0 & 0 \0 & 0 & 1 \\end{pmatrix} \begin{pmatrix}x \y \z \\end{pmatrix} = \begin{pmatrix}0 \0 \0 \\end{pmatrix}\]From this system, solve \(y = 0\) and \(x = 0\), with \(z\) being free. The eigenvector is \(\begin{pmatrix}0 \ 0 \ 1\end{pmatrix}\).For \(\lambda_2 = 1\), solve \((\mathbf{A} - \mathbf{I})\mathbf{x} = \mathbf{0}\), simplifying to:\[\begin{pmatrix}-1 & -9 & 0 \1 & -1 & 0 \0 & 0 & 0 \\end{pmatrix} \begin{pmatrix}x \y \z \\end{pmatrix} = \begin{pmatrix}0 \0 \0 \\end{pmatrix}\]From \(x = 9y\), we choose \(y = 1\): \(x = 9\), and \(z\) is free. The eigenvector is \(\begin{pmatrix}9 \ 1 \ 0\end{pmatrix}\).

Step 3: Check Algebraic and Geometric Multiplicity

The matrix is diagonalizable if the number of linearly independent eigenvectors equals the order of the matrix (3 in this case). We have:- For \(\lambda_1 = 0\): one eigenvector, when two are needed- For \(\lambda_2 = 1\): one eigenvector.Since we do not have a full set of three linearly independent eigenvectors, the matrix \(\mathbf{A}\) is **not diagonalizable**.

Key Concepts

Eigenvalues and Eigenvectors

When dealing with linear algebra, two critical components to understand are eigenvalues and eigenvectors. These help determine the properties of a matrix and have various applications.

• Eigenvalues signify by how much a transformation scales a vector in a particular direction. They can be regarded as "scaling factors" associated with a matrix.
• Eigenvectors are the non-zero vectors that change by only a scalar factor when that linear transformation is applied. Essentially, they indicate the directions of such scaling.

To find eigenvalues, we utilize the characteristic equation, which leads us to the transformation of a matrix that preserves certain vector directions. Once eigenvalues are determined, we solve for eigenvectors using equations derived from the matrix. Each eigenvalue corresponds to specific eigenvectors.

Characteristic Equation

The characteristic equation is fundamental in finding a matrix's eigenvalues. It is derived from the matrix \(\mathbf{A} - \lambda \mathbf{I}\), where \(\lambda\) represents potential eigenvalues.

• Formulating the equation: The equation is written as \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). Here, \(\mathbf{I}\) is the identity matrix, which resembles a matrix consisting of 1s on the diagonal and 0s elsewhere.

Solving this equation involves finding all values of \(\lambda\) that make the determinant of \(\mathbf{A} - \lambda \mathbf{I}\) equal to zero. This results in a polynomial whose roots are the eigenvalues of the matrix. In the example given, solving this characteristic equation yields the eigenvalues \(\lambda = 0, 0, 1\). These values play a vital role in analyzing whether a matrix is diagonalizable.

Geometric Multiplicity

Geometric multiplicity involves the number of linearly independent eigenvectors associated with an eigenvalue. It is crucial for understanding matrix diagonalization.

• Definition: Geometric multiplicity is the dimension of the subspace spanned by the eigenvectors of a particular eigenvalue. This is always less than or equal to its algebraic multiplicity (the number of times an eigenvalue appears in the characteristic equation).

In determining if a matrix is diagonalizable, it's essential that for each eigenvalue, the geometric multiplicity matches the algebraic multiplicity. Also, the total count of linearly independent eigenvectors should equal the order (size) of the matrix. In our exercise, the matrix wasn't diagonalizable due to not having enough independent eigenvectors to match each eigenvalue's multiplicity.