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

Step-by-Step Solution

Answer

The matrix \( \mathbf{A} \) is diagonalizable. The matrix \( \mathbf{P} = \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \).

Step 1: Determine the eigenvalues

To find out if the matrix \( \mathbf{A} \) is diagonalizable, we first need its eigenvalues. For matrix \( \mathbf{A} = \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix} \), find the eigenvalues by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The resulting equation will be \( \det\begin{pmatrix} -\lambda & 1 \ -1 & -\lambda \end{pmatrix} \). This simplifies to \( \lambda^2 + 1 = 0 \), yielding eigenvalues \( \lambda = i \) and \( \lambda = -i \).

Step 2: Find the eigenvectors

For each eigenvalue, find the corresponding eigenvector. Starting with \( \lambda = i \), solve \((\mathbf{A} - i\mathbf{I})\mathbf{v} = \mathbf{0}\). This results in solving \( \begin{pmatrix} -i & 1 \ -1 & -i \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\). Solving this gives the eigenvector \( \begin{pmatrix} 1 \ i \end{pmatrix} \). Similarly with \( \lambda = -i \), solve \( \begin{pmatrix} i & 1 \ -1 & i \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\) yielding eigenvector \( \begin{pmatrix} 1 \ -i \end{pmatrix} \).

Step 3: Construct the matrix \(\mathbf{P}\)

The matrix \( \mathbf{P} \) is constructed using the eigenvectors as columns, so \( \mathbf{P} = \begin{pmatrix} 1 & 1 \ i & -i \end{pmatrix} \).

Step 4: Construct the diagonal matrix \(\mathbf{D}\)

The diagonal matrix \( \mathbf{D} \) has its diagonal elements as the eigenvalues of \( \mathbf{A} \), so \( \mathbf{D} = \begin{pmatrix} i & 0 \ 0 & -i \end{pmatrix} \).

Step 5: Verify \(\mathbf{A}\) is diagonalizable

Verify the diagonalization by checking if \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \). Compute \( \mathbf{P}^{-1} \) and verify that the above matrix product equals \( \mathbf{D} \). In this case, the calculations confirm that \( \mathbf{A} \) is diagonalizable with the matrices \( \mathbf{P} \) and \( \mathbf{D} \) as obtained.

Key Concepts

Eigenvalues

Eigenvalues are a special set of scalars associated with a matrix. They are calculated from the characteristic equation, which is derived from the matrix itself. When you deduct \( \lambda \mathbf{I} \) from the matrix \( \mathbf{A} \) and then find the determinant, the zeros of this characteristic polynomial are the eigenvalues. For the given matrix \( \mathbf{A} = \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix} \), the equation becomes \( \lambda^2 + 1 = 0 \). Solving this, you will find that the eigenvalues are complex numbers: \( \lambda = i \) and \( \lambda = -i \).

It's crucial to understand that eigenvalues can be real or complex, depending on the types of numbers involved in the matrix and the characteristic equation. When a matrix has distinct eigenvalues, it is almost always diagonalizable. In some cases, complex eigenvalues are expected, especially in matrices involving rotations or certain transformations.

Knowing how to find eigenvalues is a foundational skill in linear algebra. They offer insights into the matrix's properties and behaviors, such as stability and oscillatory behavior in dynamic systems.

Eigenvectors

Eigenvectors correspond to eigenvalues and are non-zero vectors that only change by a scalar factor when a matrix is applied. To find an eigenvector, solve \((\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\). It's a system of equations derived from your matrix minus the eigenvalue times the identity matrix. For an eigenvalue \( \lambda = i \), the eigenvector \( \begin{pmatrix} 1 \ i \end{pmatrix} \) satisfies the equation.

Similarly, for \( \lambda = -i \), solve the same form of equation to get the eigenvector \( \begin{pmatrix} 1 \ -i \end{pmatrix} \). These solutions confirm that the chosen vectors remain unchanged in direction upon transformation by matrix \( \mathbf{A} \) but simply get scaled by the eigenvalue.

A matrix is diagonalizable if it has a set of linearly independent eigenvectors equal in number to the dimension of the matrix. Eigenvectors are vital in stability analysis, quantum mechanics, and other applications where transformation properties are studied. Understanding how to compute and interpret them is key in advanced mathematics and engineering.

Diagonal Matrix

A diagonal matrix is one where all off-diagonal entries are zero. In the context of diagonalizing a matrix \( \mathbf{A} \), a diagonal matrix \( \mathbf{D} \) is formed by placing eigenvalues on its diagonal. For the matrix in our problem, \( \mathbf{D} = \begin{pmatrix} i & 0 \ 0 & -i \end{pmatrix} \) with eigenvalues \( i \) and \( -i \).

When a matrix is diagonalized, it simplifies matrix operations, such as exponentiation, and makes computations more efficient. This efficiency is particularly important in practical applications where matrices model real-world systems.

To compute this, we have \( \mathbf{D} = \mathbf{P}^{-1}\mathbf{A}\mathbf{P} \), where \( \mathbf{P} \) is constructed from eigenvectors. The matrix \( \mathbf{P} \) in our case is \( \begin{pmatrix} 1 & 1 \ i & -i \end{pmatrix} \), arranged with the eigenvectors you've found earlier.

The utility of diagonal matrices is that they reveal the structure and properties of the original matrix. For instance, they can provide immediate insights into dynamics described by \( \mathbf{A} \), especially in systems where diagonalization represents a transition to normal modes.