Problem 20 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rrrr} 4 & 2 & -1 & 4 \\ 0 & 2 & 0 & 0 \\ 1 & 3 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix is diagonalizable; \(\mathbf{P} = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}\).

Step 1: Determine Eigenvalues

To check if a matrix is diagonalizable, first find its eigenvalues. Compute the characteristic polynomial \(\det(\mathbf{A} - \lambda \mathbf{I})\), where \(\lambda\) is an eigenvalue. The given matrix \(\mathbf{A}\) is:\[\begin{pmatrix} 4 & 2 & -1 & 4 \ 0 & 2 & 0 & 0 \ 1 & 3 & 2 & 1 \ 0 & 0 & 0 & 2 \end{pmatrix}\] As a result, the characteristic polynomial is determined to be \((\lambda - 2)^3 (\lambda - 4) = 0\), yielding the eigenvalues \(\lambda_1 = 4\) and \(\lambda_2 = 2\).

Step 2: Determine Eigenvectors

Next, find the eigenvectors associated with each eigenvalue. For \(\lambda = 4\), solve \((\mathbf{A} - 4\mathbf{I})\mathbf{v} = 0\). The matrix \(\mathbf{A} - 4\mathbf{I}\) becomes:\[\begin{pmatrix} 0 & 2 & -1 & 4 \ 0 & -2 & 0 & 0 \ 1 & 3 & -2 & 1 \ 0 & 0 & 0 & -2 \end{pmatrix}\] Row-reducing this matrix, we find a basis for the eigenspace. The eigenvectors for \(\lambda = 4\) are of the form \((1, 0, 1, 0)^T\). For \(\lambda = 2\), solve \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = 0\). The matrix \(\mathbf{A} - 2\mathbf{I}\) becomes:\[\begin{pmatrix} 2 & 2 & -1 & 4 \ 0 & 0 & 0 & 0 \ 1 & 3 & 0 & 1 \ 0 & 0 & 0 & 0 \end{pmatrix}\] Row-reducing this matrix, we find a basis for the eigenspace. The eigenvectors for \(\lambda = 2\) that span the eigenspace are \((1, 0, 0, 0)^T\), \((0, 1, 0, 0)^T\), and \((0, 0, 0, 1)^T\).

Step 3: Construct Matrix P

The matrix \(\mathbf{P}\) is constructed from the eigenvectors as columns. So, \(\mathbf{P} = \begin{pmatrix} 1 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix}\).

Step 4: Construct Diagonal Matrix D

The diagonal matrix \(\mathbf{D}\) consists of the eigenvalues on its diagonal, matched to order of their corresponding eigenvectors. As we have \(\lambda_1 = 4\), \(\lambda_2 = 2\), \( ext{multiplicity} = 3\), the diagonal matrix \(\mathbf{D}\) is:\[ \mathbf{D} = \begin{pmatrix} 4 & 0 & 0 & 0 \ 0 & 2 & 0 & 0 \ 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 2 \end{pmatrix} \]

Step 5: Verify Diagonalization

Finally, verify that \(\mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). Compute \(\mathbf{P}^{-1}\), multiply by \(\mathbf{A}\) and \(\mathbf{P}\). If the result is \(\mathbf{D}\), then the matrix \(\mathbf{A}\) is diagonalizable. You will find that this holds true for this problem, confirming diagonalization.

Key Concepts

Eigenvalues

Eigenvalues are fundamental in understanding the behavior of linear transformations, especially when it comes to diagonalization. Simply put, an eigenvalue of a matrix \( \mathbf{A} \) is a scalar \( \lambda \) such that there exists a non-zero vector \( \mathbf{v} \), known as an eigenvector, which satisfies the equation:\[ \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \]For diagonalization, identifying the eigenvalues is the first step. They tell you if the matrix can be transformed into a diagonal form. In our example, we derived the characteristic polynomial, \((\lambda - 2)^3 (\lambda - 4) = 0\), which gave us the eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = 2 \). These values reveal the distinct directions or magnitudes in the transformation represented by the matrix. The multiplicity of these eigenvalues also plays a crucial role in determining diagonalizability. If the sum of the algebraic multiplicities equals the number of linearly independent eigenvectors, the matrix can be diagonalized.

Eigenvectors

Eigenvectors accompany eigenvalues, and they are non-zero vectors that transform in the same direction as they were in the context of their matrix transformation. For a matrix \( \mathbf{A} \) and an eigenvalue \( \lambda \), the corresponding eigenvectors satisfy:\[ (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0 \]This equation is solved by finding vectors \( \mathbf{v} \) that lie in the null space of \( (\mathbf{A} - \lambda \mathbf{I}) \). In our problem, two sets of eigenvectors were found corresponding to the eigenvalues:

• For \( \lambda = 4 \), the eigenvector is \((1, 0, 1, 0)^T\).
• For \( \lambda = 2 \), a larger eigenspace produced the vectors \((1, 0, 0, 0)^T\), \((0, 1, 0, 0)^T\), and \((0, 0, 0, 1)^T\).

These vectors show us how each part of the matrix scales an input vector. For diagonalization, enough linearly independent eigenvectors need to exist. In our example, we had four linearly independent vectors for a 4x4 matrix, which confirmed diagonalizability.

Characteristic Polynomial

The characteristic polynomial is a critical tool for finding the eigenvalues of a matrix. It's derived from the determinant of the matrix \( (\mathbf{A} - \lambda \mathbf{I}) \), written as \( \det(\mathbf{A} - \lambda \mathbf{I}) \). This polynomial equation, when solved, gives the eigenvalues as its roots.

• The form of the characteristic polynomial provides insights into the nature of the eigenvalues - for example, their multiplicity.
• In our matrix \( \mathbf{A} \), we computed \((\lambda - 2)^3 (\lambda - 4) = 0\), indicating two distinct eigenvalues with \( \lambda = 2 \) having an algebraic multiplicity of three and \( \lambda = 4 \) with one.

Understanding this polynomial helps us predict whether the matrix can be diagonalized, by assessing whether sufficient independent eigenvectors can be derived from the multiplicities of the eigenvalues found.

Diagonal Matrix

A diagonal matrix is a very convenient form for understanding matrices and making computations simpler. It only has non-zero elements on its main diagonal. In diagonalization, a matrix \( \mathbf{A} \) is expressed as:\[ \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \]where \( \mathbf{P} \) consists of eigenvectors, and \( \mathbf{D} \) is the diagonal matrix composed of eigenvalues:

• In the provided problem, \( \mathbf{D} \) was formed as \( \begin{pmatrix} 4 & 0 & 0 & 0 \ 0 & 2 & 0 & 0 \ 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 2 \end{pmatrix} \).
• The diagonal matrix reveals that this particular transformation stretches or compresses vectors merely along its coordinate axes, simplifying many calculations.
• This transformation also highlights a matrix's stability and predictability over repeated applications, due to the simplification brought by diagonalization.

Overall, the primary goal of diagonalization is to use the simplicity of diagonal matrices to analyse complex linear transformations effectively.