Question

Let \[ A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 2 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] . Then the sum of the geometric multiplicities of the distinct eigenvalues of \( A \) is equal to ________.

Hint:

The geometric multiplicity of an eigenvalue is the dimension of its eigenspace, which is the number of linearly independent eigenvectors corresponding to that eigenvalue.

Answer 1

Correct Answer: 2

We need to find the eigenvalues of matrix \( A \). First, calculate the characteristic equation by finding the determinant of \( A - \lambda I \):
\[ \det(A - \lambda I) = \det \begin{pmatrix} 2 - \lambda & 0 & 1 \\ 1 & 2 - \lambda & 5 \\ 0 & 3 & -\lambda \\ 0 & 0 & 1 - \lambda \end{pmatrix} \] Using the cofactor expansion, we obtain the characteristic polynomial. After solving for the eigenvalues, we find the distinct eigenvalues and their geometric multiplicities. The distinct eigenvalues of \( A \) are: \[ \lambda_1 = 3, \quad \lambda_2 = 2, \quad \lambda_3 = 1 \] The geometric multiplicities are:
- For \( \lambda_1 = 3 \), the geometric multiplicity is 1.
- For \( \lambda_2 = 2 \), the geometric multiplicity is 1.
- For \( \lambda_3 = 1 \), the geometric multiplicity is 1.
Thus, the sum of the geometric multiplicities is: \[ \boxed{3} \]