Problem 16 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix \( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) and \( \mathbf{D} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \).

Step 1: Analyze the Matrix Structure

The matrix \( \mathbf{A} \) is given as \(\begin{bmatrix} 1 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{bmatrix} \). Notice that it is an upper triangular matrix whose eigenvalues are simply the elements on its diagonal.

Step 2: Determine the Eigenvalues

Since \( \mathbf{A} \) is upper triangular, its eigenvalues are the entries on the main diagonal: 1, 2, and 3.

Step 3: Find Eigenvectors for Each Eigenvalue

To be diagonalizable, the matrix must have a complete set of linearly independent eigenvectors. Find the eigenvector for each eigenvalue by solving \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0\) for \(\lambda = 1, 2, 3\).

Step 4: Solve for Eigenvectors When \(\lambda = 1\)

Set up the equation \((\mathbf{A} - 1\mathbf{I})\mathbf{v} = 0\).\[\begin{bmatrix} 0 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \]This simplification shows that \( v_2 = 0 \) and \( v_3 \) is free, so an eigenvector \( \mathbf{v}_1 \) could be \( \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \).

Step 5: Solve for Eigenvectors When \(\lambda = 2\)

Set up the equation \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = 0\).\[\begin{bmatrix} -1 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \]This simplification leads to \( v_1 = v_2 \) while \( v_3 = 0 \). An eigenvector \( \mathbf{v}_2 \) could be \( \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \).

Step 6: Solve for Eigenvectors When \(\lambda = 3\)

Set up the equation \((\mathbf{A} - 3\mathbf{I})\mathbf{v} = 0\).\[\begin{bmatrix} -2 & 1 & 0 \ 0 & -1 & 0 \ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \]This simplification shows \( v_2 = 0 \) while \( v_1 = 0 \) and \( v_3 \) is free. An eigenvector \( \mathbf{v}_3 \) could be \( \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \).

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

The matrix \( \mathbf{P} \) is formed by placing each eigenvector as columns: \[\mathbf{P} = \begin{bmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \] and the diagonal matrix \( \mathbf{D} \) with eigenvalues on the diagonal: \( \mathbf{D} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{bmatrix} \).

Step 8: Check Diagonalization

Verify that \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} = \mathbf{D} \). Calculate \( \mathbf{P}^{-1} \): \[\mathbf{P}^{-1} = \begin{bmatrix} 1 & -1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \] then \( \mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D} \). The matrix \( \mathbf{A} \) is diagonalizable.

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept when dealing with matrices, especially in the context of diagonalization. Simply put, an eigenvalue of a matrix is a special number associated with the matrix that provides insights into the matrix's properties. To find the eigenvalues of a matrix \( \mathbf{A} \), you solve the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) represents the eigenvalues and \( \mathbf{I} \) is the identity matrix of the same size as \( \mathbf{A} \).​
In this example, the given matrix \( \mathbf{A} \) is upper triangular, which means its eigenvalues are simply the numbers on the main diagonal. These are 1, 2, and 3. Having distinct eigenvalues is crucial because it influences whether a matrix is diagonalizable by assuring that there are enough independent eigenvectors. This is reflected in the matrix's eigenvalue multiplicity, which should be equal to the algebraic dimensions of its eigenspace to guarantee diagonalization. In summary, eigenvalues not only help in understanding the intrinsic behavior of matrices but also play a pivotal role in determining the diagonalizability of matrices.

Eigenvectors

Eigenvectors go hand in hand with eigenvalues. For each eigenvalue of a matrix, there is at least one corresponding eigenvector. An eigenvector of a matrix \( \mathbf{A} \) associated with an eigenvalue \( \lambda \) is a non-zero vector \( \mathbf{v} \) that satisfies the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \).​
In the context of our exercise, the given matrix \( \mathbf{A} \) has eigenvectors corresponding to its eigenvalues 1, 2, and 3. Specifically:

• For \( \lambda = 1 \), a valid eigenvector is \( \begin{bmatrix} 1 \, 0 \, 0 \end{bmatrix} \).
• For \( \lambda = 2 \), a valid eigenvector is \( \begin{bmatrix} 1 \, 1 \, 0 \end{bmatrix} \).
• For \( \lambda = 3 \), a valid eigenvector is \( \begin{bmatrix} 0 \, 0 \, 1 \end{bmatrix} \).

Finding these vectors is essential because they form the basis that, when placed as columns in a transformation matrix \( \mathbf{P} \), diagonalizes \( \mathbf{A} \). Each eigenvector characterizes one direction in which a linear transformation with the matrix is scaled by its respective eigenvalue. This central concept emphasizes that having a sufficient number of linearly independent eigenvectors confirms matrix diagonalizability.

Matrix Algebra

Matrix Algebra is the framework within which we perform operations on matrices, such as addition, multiplication, and finding inverses. Understanding these operations provides the tools necessary for solving systems of linear equations, transforming geometric objects, and more. When working with matrices like in our diagonalization exercise, we leverage matrix multiplication and the concept of matrix inverses.
In diagonalization, you often need to calculate the inverse of the matrix \( \mathbf{P} \), which is used to transform the original matrix \( \mathbf{A} \) into a diagonal matrix \( \mathbf{D} \). This is done by ensuring that \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \). The step of finding \( \mathbf{P}^{-1} \), and verifying the equation embodies the heart of matrix algebra, where computational precision is key.
To fully utilize matrix algebra, consider:

• Matrix Multiplication: Enables the transformation of matrices and is not commutative, meaning \( \mathbf{A} \mathbf{B} eq \mathbf{B} \mathbf{A} \).
• Determining Inverses: Helpful for reversible transformations; a matrix must be square and non-singular.
• Identity Matrix: Plays a pivotal role in maintaining the original value during multiplication, like the number 1 in arithmetic operations.

These fundamental operations are indispensable in computing solutions to linear algebra problems and concepts like diagonalization.

Linear Independence

Linear Independence is a critical concept in linear algebra that refers to a set of vectors not being linearly expressible by any other combination of vectors from this set. Being linearly independent assures that no vector in the set can be represented as a scalar multiple or a linear combination of others in the set.
In the context of diagonalization, having a complete set of linearly independent eigenvectors is crucial. For the matrix \( \mathbf{A} \) to be diagonalizable, the eigenvectors determined need to be linearly independent. This means that in each eigenspace (the space of all eigenvectors associated with a particular eigenvalue), the set of eigenvectors spans the space without redundancy.
To check the linear independence of eigenvectors obtained from the example:

• Eigenvectors \( \begin{bmatrix} 1 \, 0 \, 0 \end{bmatrix} \), \( \begin{bmatrix} 1 \, 1 \, 0 \end{bmatrix} \), and \( \begin{bmatrix} 0 \, 0 \, 1 \end{bmatrix} \) should not be expressible as linear combinations of each other.

If they are independent, they form a basis for \( \mathbf{R}^3 \), confirming that the matrix \( \mathbf{A} \) can indeed be diagonalized using these eigenvectors to form \( \mathbf{P} \). This guarantees that there is a unique linear transformation path that scales these directions precisely, underscoring the significance of linear independence in transformation matrices.