1. **Problem Statement:** Given a 3x3 matrix $A$ with characteristic polynomial $$P(\lambda) = (\lambda - 3)(\lambda + 1)^2,$$ we need to: a) List all eigenvalues with their algebraic multiplicities. b) Describe what additional information is needed to determine the geometric multiplicity of each eigenvalue. c) Give an example when $A$ would not be diagonalizable. 2. **Step a: Eigenvalues and Algebraic Multiplicities** - The eigenvalues are the roots of the characteristic polynomial. - From $$P(\lambda) = (\lambda - 3)(\lambda + 1)^2,$$ the eigenvalues are: - $\lambda_1 = 3$ with algebraic multiplicity 1 (since $(\lambda - 3)$ appears once). - $\lambda_2 = -1$ with algebraic multiplicity 2 (since $(\lambda + 1)^2$ appears squared). 3. **Step b: Geometric Multiplicity** - The geometric multiplicity of an eigenvalue is the dimension of its eigenspace, i.e., the number of linearly independent eigenvectors associated with it. - To determine geometric multiplicity, we need to know the nullity of $(A - \lambda I)$ for each eigenvalue $\lambda$. - This requires information about the matrix $A$ itself or its eigenspaces, not just the characteristic polynomial. - In other words, algebraic multiplicity is given by the polynomial, but geometric multiplicity depends on the matrix's structure. 4. **Step c: When is $A$ not diagonalizable?** - A matrix is diagonalizable if and only if, for each eigenvalue, the geometric multiplicity equals the algebraic multiplicity. - Since $\lambda = -1$ has algebraic multiplicity 2, if its geometric multiplicity is less than 2 (i.e., only 1 independent eigenvector), then $A$ is not diagonalizable. - Example: If the eigenspace for $\lambda = -1$ is one-dimensional, $A$ is not diagonalizable. **Final answers:** - a) Eigenvalues: $3$ (algebraic multiplicity 1), $-1$ (algebraic multiplicity 2). - b) Need the dimension of eigenspaces (geometric multiplicities) from $(A - \lambda I)$ nullities. - c) $A$ is not diagonalizable if geometric multiplicity of $-1$ is less than 2.