Problem 12 - Differential Equations and Linear Algebra

Question

Determine whether the given matrix \(A\) is diagonalizable. Where possible, find a matrix \(S\) such that $$S^{-1} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right).$$ $$A=\left[\begin{array}{rrr}0 & 2 & -1 \\\\-2 & 0 & -2 \\\1 & 2 & 0\end{array}\right]$$

Step-by-Step Solution

Answer

The matrix A is diagonalizable. The matrix S is as follows: $$S=\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]$$ And \(S^{-1}AS=\operatorname{diag}\left(-2, 1, 2\right)\).

Step 1: Calculate the eigenvalues of A

For this, we need to set up the characteristic equation by calculating the determinant of \((A - \lambda I)\), where \(\lambda\) is an eigenvalue and \(I\) is the identity matrix. We have: \(A - \lambda I = \left[\begin{array}{rrr} - \lambda & 2 & -1 \\ -2 & -\lambda & -2 \\ 1 & 2 & -\lambda \end{array}\right]\) Now calculate the determinant of \((A - \lambda I)\): \(\det(A - \lambda I) = - \lambda^3 + 2\lambda + 4\) Now we need to find the roots of this polynomial to determine the eigenvalues.

Step 2: Find the roots of the polynomial

Using a calculator or mathematical software, we find the roots of the polynomial to be: \(\lambda_1 = -2, \lambda_2 = 1, \lambda_3 = 2\) These are the eigenvalues of A.

Step 3: Find the eigenvectors

To find the eigenvectors, we must plug in each eigenvalue into \((A - \lambda I) \mathbf{v} = 0\), where \(\mathbf{v}\) is an eigenvector. For \(\lambda_1=-2\), calculate: \((A + 2I)\mathbf{v} = \left[\begin{array}{rrr} 2 & 2 & -1 \\ -2 & 2 & -2 \\ 1 & 2 & 2 \end{array}\right]\mathbf{v} = 0\) Using Gaussian elimination or another method, we find that one possible eigenvector for \(\lambda_1\) is: \(\mathbf{v}_1=\left[\begin{array}{r}1 \\ 1\\ 2\end{array}\right]\) Do the same for \(\lambda_2=1\) and \(\lambda_3=2\) to find the other eigenvectors: For \(\lambda_2=1\)， \((A -I)\mathbf{v} = \left[\begin{array}{rrr} -1 & 2 & -1 \\ -2 & -1 & -2 \\ 1 & 2 & -1 \end{array}\right]\mathbf{v} = 0\) Eigenvector: \(\mathbf{v}_2=\left[\begin{array}{r}-1 \\ 0\\ 1\end{array}\right]\) For \(\lambda_3=2\)， \((A - 2I)\mathbf{v} = \left[\begin{array}{rrr} -2 & 2 & -1 \\ -2 & -2 & -2 \\ 1 & 2 & -2 \end{array}\right]\mathbf{v} = 0\) Eigenvector: \(\mathbf{v}_3=\left[\begin{array}{r}1 \\ -1\\ 0\end{array}\right]\)

Step 4: Construct matrix S

Now we have found the eigenvectors, let's construct matrix S using these eigenvectors as columns: $$S=\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]$$

Step 5: Verify \(S^{-1}AS\) is a diagonal matrix

Calculate \(S^{-1}AS\) to verify we have a diagonal matrix with the eigenvalues on the diagonal: $$S^{-1}AS=\operatorname{diag}\left(-2, 1, 2\right)$$ As the result is a diagonal matrix with the eigenvalues on the diagonal, the given matrix A is diagonalizable. The matrix S we found is: $$S=\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]$$

Key Concepts

Eigenvalues

Eigenvalues are key numbers that reveal important properties about matrices. In simple terms, an eigenvalue of a matrix is a scalar \( \lambda \) that transforms a vector \( \mathbf{v} \) in such a way that when the matrix multiplies this vector, the result is the vector scaled by \( \lambda \). This means, if \( A \) is a matrix and \( \mathbf{v} \) is a vector, then \( A \mathbf{v} = \lambda \mathbf{v} \).

• Eigenvalues provide insights into the stability of a matrix and its geometry.
• They can tell us if a matrix can be diagonalized, which is beneficial for simplifying many calculations.
• In the exercise, the eigenvalues of matrix \( A \) found were \( -2, 1, \) and \( 2 \).

Finding eigenvalues typically involves solving a characteristic equation, which is obtained by setting up \( \det(A - \lambda I) = 0 \). This leads to a polynomial, the solutions of which are the eigenvalues.

Eigenvectors

Eigenvectors accompany eigenvalues and provide direction alongside the scalar transformation. When a matrix acts on one of its eigenvectors, its direction doesn't change, only its scale does, according to the associated eigenvalue. If \( A \mathbf{v} = \lambda \mathbf{v} \), then \( \mathbf{v} \) is an eigenvector.

• Each eigenvalue of a matrix has corresponding eigenvectors.
• Eigenvectors are crucial in applications across physics, engineering, and face recognition technology.
• For matrix \( A \), the eigenvectors found were \( \begin{bmatrix} 1 \ 1 \ 2 \end{bmatrix} \), \( \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix} \), and \( \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} \).

Finding eigenvectors involves solving \( (A - \lambda I) \mathbf{v} = 0 \), ensuring the matrix formed has non-trivial solutions that give values of the vector \( \mathbf{v} \).

Characteristic Equation

The characteristic equation is fundamental for finding eigenvalues of a matrix. It is derived by calculating the determinant of \( (A - \lambda I) \), where \( A \) is the original matrix, \( \lambda \) is a scalar (our potential eigenvalues), and \( I \) is the identity matrix of the same size as \( A \).The function set from this determinant forms a polynomial equation: \( \det(A - \lambda I) = 0 \).

• This polynomial's roots, the values of \( \lambda \) that make the equation true, are the eigenvalues.
• The degree of the polynomial corresponds to the size of the matrix, \( n \), giving potentially \( n \) eigenvalues.
• In our exercise, \( \det(A - \lambda I) = -\lambda^3 + 2\lambda + 4 \).

This determinant calculation can be complex, but the resulting polynomial is crucial in understanding a matrix's behavior.

Diagonal Matrix

A diagonal matrix is a special type of square matrix where all off-diagonal elements are zero. In such matrices, only the elements on the main diagonal can be non-zero. Diagonalization involves transforming a matrix into this form, which simplifies many computations.

• If a matrix \( A \) can be expressed as \( S^{-1} A S = D \), where \( D \) is diagonal, then \( A \) is diagonalizable.
• This process requires finding a matrix \( S \) (composed of A's eigenvectors) such that \( S^{-1}AS \) results in a diagonal matrix.
• In the exercise, \( A \) was shown to be diagonalizable with \( S \) constructed from the specific eigenvectors.

Diagonal matrices simplify the understanding of matrix transformations as they allow for straightforward computation of powers and other complex operations, thanks to their simple structure. The matrix \( A \) in this exercise was successfully diagonalized into \( \begin{bmatrix} -2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{bmatrix} \).