1. **Problem Statement:** A linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ has eigenvalues (Băzel numbers) 1, 2, and 3. A) Find the possible geometric multiplicities of the eigenvalues. B) Determine under what condition $T$ would fail to be diagonalizable. 2. **Key Concepts and Formulas:** - The **algebraic multiplicity** of an eigenvalue is its multiplicity as a root of the characteristic polynomial. - The **geometric multiplicity** of an eigenvalue is the dimension of its eigenspace (number of linearly independent eigenvectors). - For each eigenvalue $\lambda$, geometric multiplicity $\leq$ algebraic multiplicity. - A matrix (or linear transformation) is diagonalizable if and only if the sum of geometric multiplicities equals the dimension of the space (here 3). 3. **Step A: Possible Geometric Multiplicities** - Since eigenvalues are distinct (1, 2, 3), each has algebraic multiplicity 1. - Therefore, geometric multiplicity for each eigenvalue is exactly 1. - So, possible geometric multiplicities are: $$ m_g(1) = 1, \quad m_g(2) = 1, \quad m_g(3) = 1 $$ 4. **Step B: Condition for $T$ to fail diagonalizability** - $T$ fails to be diagonalizable if the sum of geometric multiplicities is less than 3. - This can only happen if at least one eigenvalue has geometric multiplicity less than its algebraic multiplicity. - Since all eigenvalues are distinct with algebraic multiplicity 1, geometric multiplicity cannot be less than 1. - Hence, $T$ is always diagonalizable in this case. - If eigenvalues were repeated, failure occurs if geometric multiplicity $<$ algebraic multiplicity for any eigenvalue. **Final answers:** A) Geometric multiplicities are all 1 for eigenvalues 1, 2, and 3. B) $T$ fails to be diagonalizable if any eigenvalue's geometric multiplicity is less than its algebraic multiplicity, which cannot happen here since eigenvalues are distinct.