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From eigenvalues to eigentheory: the two multiplicities

Vol I taught you to find eigenvalues and eigenvectors. Now we ask the deeper question: how many independent eigenvectors does an eigenvalue actually have, and how does that number compare to how often it appears as a root?

Where Volume I left off

In Volume I you computed an eigenvalue lambda as a root of the characteristic polynomial det(A - lambda I), then solved (A - lambda I)v = 0 for an eigenvector. That recipe works, but it hides a subtlety: a single lambda can have many independent eigenvectors, or surprisingly few. Eigentheory is the study of exactly how many, and what that count tells you about the operator.

Setting lambda's root structure aside for a moment, the cleanest object to track is not the eigenvalue but the whole space of vectors it fixes-up-to-scaling. That space is the eigenspace, and packaging eigenvectors into a subspace is the move that turns ad hoc computation into a theory.

The eigenspace as a kernel

Fix an eigenvalue lambda. The eigenspace E_lambda is exactly the null space of (A - lambda I): every nonzero vector there is an eigenvector, and including the zero vector makes it a genuine subspace. So eigenspaces inherit everything you know about kernels — dimension, bases, the rank-nullity theorem.

A = [4, 1; 0, 4]            # 2x2, lambda = 4 is the only eigenvalue

characteristic poly: det(A - x I) = (4 - x)^2
  -> lambda = 4 appears with multiplicity 2 as a ROOT

eigenspace E_4 = null(A - 4I) = null([0, 1; 0, 0])
  (A - 4I) v = 0  =>  v = (t, 0)
  basis: { (1, 0) }      -> dim E_4 = 1

So: root appears twice, but only 1 independent eigenvector.
A 2x2 where the eigenvalue is a double root yet the eigenspace is one-dimensional.

Algebraic vs geometric multiplicity

Two different counts attach to each lambda. The algebraic multiplicity is how many times (x - lambda) divides the characteristic polynomial — its multiplicity as a root. The geometric multiplicity is dim E_lambda — how many independent eigenvectors you actually get. The example above has algebraic 2 but geometric 1.

  1. Compute the characteristic equation det(A - x I) = 0 and factor it; the exponent on each (x - lambda) is that eigenvalue's algebraic multiplicity.
  2. For each lambda, row-reduce (A - lambda I) and count free variables; that count is dim E_lambda, the geometric multiplicity.
  3. Compare the two for every eigenvalue. Any gap (geometric < algebraic) flags trouble for diagonalization, which Guide 3 turns into a precise criterion.