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

Step-by-Step Solution

Answer

\(\mathbf{A}\) is diagonalizable with \(\mathbf{P} = \begin{pmatrix} 1 & 0 & 9 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\).

Step 1: Identify the Matrix

The matrix given is \( \mathbf{A} = \begin{pmatrix} 0 & -9 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} \). We need to determine if \( \mathbf{A} \) is diagonalizable.

Step 2: Find Characteristic Polynomial

Calculate the characteristic polynomial by solving \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This gives: \[ \mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} -\lambda & -9 & 0 \ 1 & -\lambda & 0 \ 0 & 0 & 1-\lambda \end{pmatrix} \]. The determinant is: \(-\lambda((1-\lambda)(-\lambda) - (-9 \cdot 0)) = 0\), and simplifying we get \(\lambda^3 - \lambda^2 = 0\).

Step 3: Solve Characteristic Equation

Factor the characteristic polynomial \( \lambda^2(\lambda - 1) = 0 \). The roots are \( \lambda_1 = 0 \) with a multiplicity of 2, and \( \lambda_2 = 1 \).

Step 4: Determine Eigenvectors for \(\lambda = 0\)

For \( \lambda = 0 \), solve \((\mathbf{A} - 0 \cdot \mathbf{I})\mathbf{x} = 0\). This gives the system: \( -9y = 0 \), \( x = 0 \). The eigenvector is \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \). However, since the algebraic and geometric multiplicity need to match for diagonalizability, check for another vector, finding another eigenvector \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \).

Step 5: Determine Eigenvectors for \(\lambda = 1\)

For \( \lambda = 1 \), solve \((\mathbf{A} - 1 \mathbf{I})\mathbf{x} = 0\). The equation becomes \( \begin{pmatrix} -1 & -9 & 0 \ 1 & -1 & 0 \ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \). This determines the eigenvector \( \begin{pmatrix} 9y \ 9 \ z \end{pmatrix} \), or simplified \( \begin{pmatrix} 9 \ 1 \ 0 \end{pmatrix} \).

Step 6: Verify Linearly Independent Eigenvectors

The eigenvectors are \( \mathbf{v_1} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \), \( \mathbf{v_2} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \), and \( \mathbf{v_3} = \begin{pmatrix} 9 \ 1 \ 0 \end{pmatrix} \). Check that they are linearly independent.

Step 7: Construct \(\mathbf{P}\) and \(\mathbf{D}\)

With linearly independent eigenvectors, form \( \mathbf{P} \) using these vectors as columns: \( \mathbf{P} = \begin{pmatrix} 1 & 0 & 9 \ 0 & 1 & 1 \ 0 & 0 & 0 \end{pmatrix} \). Place eigenvalues on the diagonal of \( \mathbf{D} \): \( \mathbf{D} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} \).

Step 8: Conclusion

\( \mathbf{A} \) is diagonalizable since there are enough linearly independent eigenvectors (3). The matrix \( \mathbf{P} \) is as found and \( \mathbf{D} \) has the eigenvalues.

Key Concepts

Eigenvalues

Eigenvalues are special numbers associated with a matrix that provide insight into its properties, such as whether it can be diagonalized. To discover them, we solve the characteristic equation. This involves finding the determinant of the matrix \( \mathbf{A} - \lambda \mathbf{I} \), where \( \lambda \) are the eigenvalues we need.

• In our exercise, the characteristic polynomial resulted in \( \lambda^3 - \lambda^2 = 0 \).
• This polynomial can be factored to identify the eigenvalues as \( \lambda_1 = 0 \) (with multiplicity of 2) and \( \lambda_2 = 1 \).

Finding eigenvalues helps us determine the structure of the matrix, and is a crucial step towards diagonalization.

Eigenvectors

Eigenvectors are vectors that indicate directions in which a linear transformation acts by stretching or shivining, without rotating. They are associated with specific eigenvalues. To find them, solve the equation:\( (\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0} \).

• For \( \lambda = 0 \), the eigenvectors are \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \) and \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \).
• For \( \lambda = 1 \), the eigenvector is \( \begin{pmatrix} 9 \ 1 \ 0 \end{pmatrix} \).

These vectors form a basis for the subspace associated with their eigenvalue, allowing for the construction of the essential matrices for diagonalization.

Characteristic Polynomial

The characteristic polynomial is a crucial tool in linear algebra, representing how matrix transformation behaves. It is obtained when we compute the determinant of the matrix \( \mathbf{A} - \lambda \mathbf{I} \). For our exercise:

• The characteristic polynomial derived was \( \lambda^3 - \lambda^2 = 0 \).
• Factoring gives \( \lambda^2(\lambda - 1) = 0 \), providing our eigenvalues.

This polynomial encapsulates information about the eigenvalues, and each term's degree indicates the potential multiplicity of corresponding eigenvalues. Understanding it is fundamental for steps like determining matrix diagonalizability.

Algebraic Multiplicity

Algebraic multiplicity refers to how many times an eigenvalue appears as a root in the characteristic polynomial. It tells us how many times a specific eigenvalue contributes to the count of the matrix's characteristic equation.

• For eigenvalue \( \lambda_1 = 0 \), the algebraic multiplicity is 2.
• For eigenvalue \( \lambda_2 = 1 \), the multiplicity is 1.

This concept is important because it impacts diagonalizability. A matrix is diagonalizable if the sum of eigenvectors for each distinct eigenvalue equals the dimension of the matrix – a condition enriched by the algebraic multiplicity.

Geometric Multiplicity

Geometric multiplicity measures the number of linearly independent eigenvectors associated with an eigenvalue. It tells us about the size of the eigenvector space for each eigenvalue.

• For \( \lambda_1 = 0 \), the geometric multiplicity is 2, matching its algebraic multiplicity.
• For \( \lambda_2 = 1 \), it is 1.

A key requirement for a matrix to be diagonalizable is that for each eigenvalue, algebraic multiplicity must not exceed geometric multiplicity. This ensures a sufficient number of eigenvectors are available, confirming the "freeness" from dependency needed for matrix diagonalization.