1. **Problem statement:** Given a 3x3 matrix $A$ with characteristic polynomial $$P(\lambda) = (\lambda - 3)(\lambda + 1)^2,$$ answer the following: 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 of when $A$ would not be diagonalizable. 2. **Recall definitions:** - The **eigenvalues** of $A$ are the roots of the characteristic polynomial. - The **algebraic multiplicity** of an eigenvalue is its multiplicity as a root of $P(\lambda)$. - The **geometric multiplicity** of an eigenvalue is the dimension of its eigenspace (number of linearly independent eigenvectors). - A matrix is **diagonalizable** if the sum of geometric multiplicities equals the size of the matrix. 3. **Part (a): Find eigenvalues and algebraic multiplicities** - 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) 4. **Part (b): Additional information for geometric multiplicity** - To find geometric multiplicity, we need to know the dimension of the null space of $(A - \lambda I)$ for each eigenvalue $\lambda$. - This requires knowledge of the matrix $A$ itself or its eigenspaces. - Specifically, we need to find the rank or nullity of $(A - 3I)$ and $(A + I)$. 5. **Part (c): Example when $A$ is not diagonalizable** - If the geometric multiplicity of $\lambda = -1$ is less than its algebraic multiplicity 2, i.e., if there is only one linearly independent eigenvector for $\lambda = -1$, then $A$ is not diagonalizable. - For example, if the eigenspace for $\lambda = -1$ is one-dimensional, $A$ has a Jordan block of size 2 for $\lambda = -1$. **Final answers:** - a) Eigenvalues: $3$ (algebraic multiplicity 1), $-1$ (algebraic multiplicity 2). - b) Need the dimension of eigenspaces (nullity of $(A - \lambda I)$) to determine geometric multiplicities. - c) $A$ is not diagonalizable if geometric multiplicity of $-1$ is less than 2, e.g., only one eigenvector for $-1$.