Problem 33 - Differential Equations and Linear Algebra

Question

Use Jordan canonical forms to determine whether the given pair of matrices are similar. \(A=\left[\begin{array}{rrr}7 & 1 & 0 \\ -1 & 5 & 0 \\ 1 & 0 & 6\end{array}\right] ; B=\left[\begin{array}{rrr}6 & -1 & 1 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]\).

Step-by-Step Solution

Answer

The given pair of matrices A and B are not similar, as their Jordan canonical forms are different: For matrix A, the Jordan canonical form is: \( J_A = \begin{bmatrix} 7 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{bmatrix} \) For matrix B, the Jordan canonical form is: \( J_B = \begin{bmatrix} 6 & 1 & 0 \\ 0 & 6 & 1 \\ 0 & 0 & 6 \end{bmatrix} \)

Step 1: Find the characteristic polynomial of the matrices

First, we will find the characteristic polynomial of the matrices A and B using the equation det(A - λI) and det(B - λI) respectively, where λ represents the eigenvalues and I is the identity matrix. For matrix A: \( det(A - \lambda I)= \left | \begin{array}{ccc} 7-\lambda & 1 & 0 \\ -1 & 5-\lambda & 0 \\ 1 & 0 & 6-\lambda \end{array} \right | = (7-\lambda)((5-\lambda)(6-\lambda)) \) For matrix B: \( det(B - \lambda I) = \left | \begin{array}{ccc} 6 - \lambda & -1 & 1 \\ 0 & 6 - \lambda & 0 \\ 0 & 0 & 6 - \lambda \end{array} \right | = (6-\lambda)^3 \)

Step 2: Compute the eigenvalues of the matrices

Now, we will compute the eigenvalues of the matrices A and B by solving the characteristic polynomial equations. For matrix A: \( (7-\lambda)((5-\lambda)(6-\lambda)) = 0 \) This gives us the eigenvalues λ = 7, 5, and 6. For matrix B: \( (6-\lambda)^3 = 0 \) This gives us the eigenvalue λ = 6 with algebraic multiplicity 3.

Step 3: Calculate the generalized eigenvectors for each eigenvalue

We will now find the generalized eigenvectors for each eigenvalue of the given matrices. For matrix A, the eigenvectors for each eigenvalue are: λ = 7: \((A-7I)v_1 = 0 \rightarrow v_1 = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\) λ = 5: \((A-5I)v_2 = 0 \rightarrow v_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}\) λ = 6: \((A-6I)v_3 = 0 \rightarrow v_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\) For matrix B, since we have only one eigenvalue λ = 6 with algebraic multiplicity 3, we need to find 3 generalized eigenvectors: \( (B - 6I)v_1 = 0 \rightarrow v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \) \( (B - 6I)^2v_2 = 0 \rightarrow v_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \) \( (B - 6I)^3v_3 = 0 \rightarrow v_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \)

Step 4: Construct the Jordan canonical form

Finally, we will construct the Jordan canonical form for matrices A and B using the obtained generalized eigenvectors. For matrix A, the Jordan canonical form is: \( J_A = \begin{bmatrix} 7 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{bmatrix} \) For matrix B, the Jordan canonical form is: \( J_B = \begin{bmatrix} 6 & 1 & 0 \\ 0 & 6 & 1 \\ 0 & 0 & 6 \end{bmatrix} \) Since the Jordan canonical forms of matrix A and matrix B are not the same, the given pair of matrices are not similar.

Key Concepts

Matrix Similarity

Matrix similarity is a concept in linear algebra where two matrices are considered similar if one can be transformed into the other through a change of basis. Specifically, matrices A and B are similar if there exists an invertible matrix P such that \( A = PBP^{-1} \). Two similar matrices represent the same linear transformation but with respect to different bases. When matrices are similar, they have the same eigenvalues and the same number of eigenvectors, making them crucially important in understanding the behavior of linear transformations.

The existence of a Jordan canonical form allows us to simplify matrices into a form that is easier to work with regarding similarity. If two matrices have different Jordan canonical forms, they are not similar. In the given exercise, matrices A and B were found to have different Jordan canonical forms, which shows they are not similar, underscoring the utility of this technique.

Eigenvalues

Eigenvalues are scalars that provide important information about a matrix's behavior, particularly in operations involving vectors. They can be thought of as the factors by which a corresponding eigenvector is scaled during a linear transformation. Mathematically, an eigenvalue \( \lambda \) of a matrix A is a scalar such that there is a non-zero vector v, the eigenvector, satisfying \( Av = \lambda v \).

To find eigenvalues, one must solve the characteristic polynomial equation \( \text{det}(A - \lambda I) = 0 \). For the matrices in the exercise, matrix A has eigenvalues 7, 5, and 6, while matrix B has a repeated eigenvalue of 6 (with an algebraic multiplicity of 3). This difference in eigenvalues immediately indicates that the matrices cannot be similar.

Generalized Eigenvectors

In situations where a matrix does not have enough linearly independent eigenvectors to form a basis, generalized eigenvectors become necessary. They complement the usual eigenvectors to complete a basis for the space. Generalized eigenvectors satisfy the equation \( (A - \lambda I)^k v = 0 \) for \( k > 1 \), filling gaps when the algebraic multiplicity exceeds the geometric multiplicity.

In the solved problem, matrix B has an eigenvalue with a higher algebraic multiplicity than its geometric multiplicity, requiring generalized eigenvectors to accurately represent its behavior. This condition leads to additional Jordan blocks in its canonical form, another indicator of non-similarity with matrix A, which has distinct eigenvectors corresponding to each of its distinct eigenvalues.

Algebraic Multiplicity

Algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the matrix's characteristic polynomial. It indicates how many times an eigenvalue is repeated in the spectrum of the matrix. Importantly, algebraic multiplicity is related to, but different from, geometric multiplicity, which counts the number of independent eigenvectors associated with an eigenvalue.

In the exercise, matrix B has the eigenvalue 6 with an algebraic multiplicity of 3, whereas matrix A has all distinct eigenvalues, each with an algebraic multiplicity of 1. This discrepancy in algebraic multiplicity further solidifies the understanding that the matrices are not similar, as similar matrices must have matching lists of eigenvalues, including their multiplicities.