Question

The characteristic polynomial of the matrix is the matrix has two real eigenvalues . (a) find these eigenvalues, their algebraic multiplicities (am) , and dimensions of the corresponding eigenspaces (gm). has algebraic multiplicity (am) . the dimension of the corresponding eigenspace (gm) is . has algebraic multiplicity (am) . the dimension of the corresponding eigenspace (gm) is . (b) is the matrix defective? check the true statements below:

a. the matrix is not defective because for at least one eigenvalue .
b. the matrix is defective because for at least one of the eigenvaues, .
c. the matrix is defective because it only has two eigenvalues
d. the matrix is not defective because it has four eigenvalues if counted with their multiplicity

Answer 1

In order to find the eigenvalues, we need to solve the characteristic polynomial of the matrix. The characteristic polynomial is found by setting the determinant of the matrix minus the identity matrix equal to zero.

The correct answer is statement a): The matrix is not defective because for at least one eigenvalue.

Step-by-step explanation:

Let's assume the matrix is A. So, we need to solve the equation |A - λI| = 0, where λ is the eigenvalue and I is the identity matrix.

Once we find the eigenvalues, we can determine their algebraic multiplicities by looking at the power of the eigenvalue in the characteristic polynomial.

The algebraic multiplicity is the exponent to which the eigenvalue is raised.

The dimension of the corresponding eigenspace is the number of linearly independent eigenvectors associated with that eigenvalue. We can find the dimension of the eigenspace by solving the equation (A - λI)v = 0, where v represents the eigenvector.

In order to answer part (b) about whether the matrix is defective or not, we need to check the statements provided. We can determine if the matrix is defective by comparing the dimensions of the eigenspaces to their algebraic multiplicities.

1. Statement a is true. If we have at least one eigenvalue with algebraic multiplicity equal to its dimension, then the matrix is not defective.
2. Statement b is false. The matrix is not defective if statement a is true.
3. Statement c is false. The matrix can have any number of eigenvalues, and it does not determine if it is defective or not.
4. Statement d is true. The number of eigenvalues counted with their multiplicities should equal the size of the matrix. If it does, then the matrix is not defective.

Therefore, the correct answer is statement a: The matrix is not defective because for at least one eigenvalue, the algebraic multiplicity equals the dimension of the eigenspace.