Determine if having an eigenvalue guarantees a generalized eigenspace decomposition across different field types (real and complex vector spaces). | Step-by-Step Solution

What We're Solving:

We need to determine whether having at least one eigenvalue is enough to guarantee that a linear transformation can be decomposed into generalized eigenspaces. This is a deep question about the structure of linear operators!

The Approach:

This is essentially asking about the relationship between eigenvalues and the Jordan canonical form. We'll explore this by examining what conditions are truly necessary for generalized eigenspace decomposition, and we'll see how the field we're working over (real vs. complex numbers) makes a crucial difference!

Step-by-Step Solution:

Step 1: Understanding Generalized Eigenspace Decomposition A generalized eigenspace decomposition means we can write our vector space V as: V = G(λ₁) ⊕ G(λ₂) ⊕ ... ⊕ G(λₖ) where G(λᵢ) is the generalized eigenspace for eigenvalue λᵢ.

Step 2: The Key Insight - What We Really Need For this decomposition to work, we need the minimal polynomial of our operator to split completely into linear factors. Having just one eigenvalue isn't enough - we need ALL roots of the minimal polynomial to be eigenvalues!

Step 3: Complex Vector Spaces - The Good News! Over ℂ (complex numbers): Every polynomial splits completely into linear factors by the Fundamental Theorem of Algebra. So if T: V → V is any linear operator on a finite-dimensional complex vector space, then YES - having eigenvalues (which always exist) guarantees generalized eigenspace decomposition.

Step 4: Real Vector Spaces - Where Things Get Tricky Over ℝ (real numbers): Polynomials don't always split completely. Consider the rotation matrix: T = [cos θ -sin θ] [sin θ cos θ] for θ ≠ 0, π

This has minimal polynomial x² - 2cos(θ)x + 1, which has complex roots e^(iθ) and e^(-iθ). Since these aren't real, T has NO real eigenvalues, so generalized eigenspace decomposition over ℝ is impossible.

Step 5: The Sufficient Condition The correct sufficient condition is: "All roots of the minimal polynomial lie in the field we're working over."

The Answer:

• Over ℂ: YES - having eigenvalues is sufficient (and eigenvalues always exist)
• Over ℝ: NO - having some eigenvalues isn't enough; we need ALL roots of the minimal polynomial to be real
• General principle: We need the minimal polynomial to split completely over our field

Memory Tip:

Think "Complete Complex" - over the complex numbers, polynomials split completely, so generalized eigenspace decomposition always works. Over the reals, some polynomials are "stubborn" and won't factor nicely, blocking the decomposition!

Great question - this really gets to the heart of why we often prefer working over algebraically closed fields like ℂ in linear algebra!