260 CHAPTER 12. SPECTRAL THEORY

Corollary 12.1.19 Let A be an n× n matrix and suppose the characteristic polynomialfactors completely and that the eigenvalues are λ 1, · · · ,λ m. If no eigenvalue is defective,then A is nondefective. If some eigenvalue is defective, then A is defective.

Proof: Let{

vij

}ri

j=1≡ β i be a basis for the eigenspace of λ i. First I claim that

{β 1, · · · ,β m}

is linearly independent. To see this, suppose wi ∈ span(β i) and ∑mi=1 wi = 0. Then by The-

orem 12.1.16, each wi = 0 since otherwise, you would have a nontrivial linear combinationof eigenvectors associated with distinct eigenvalues which is 0. Now suppose

∑i

∑j

cijv

ij = 0

Letting wi = ∑ j cijvi

j, it follows from what was just observed that each wi = 0. Now theindependence of the vectors in β i implies that for each i,ci

j = 0 for each j. Thus thesevectors {β 1, · · · ,β m} are linearly independent as claimed. Letting |β i| denote the dimen-sion of span(β i) , which equals the geometric multiplicity of λ i and letting mi denote thealgebraic multiplicity of λ i it follows that

∑i|β i| ≤ n = ∑

imi

If each |β i| = mi, then A is nondefective because {β 1, · · · ,β m} will then be a basis ofeigenvectors. This is the case of no defective eigenvalues.

In case |β i|< mi for some i, then A must be defective because if not, you would have abasis of eigenvectors and you could let γ j be those which pertain to λ j. Then γ j would be

an independent set of vectors and therefore, the number of vectors in γ j denoted as∣∣∣γ j

∣∣∣ is

no more than∣∣∣β j

∣∣∣. But then,

∑j

∣∣∣γ j

∣∣∣≤∑j

∣∣∣β j

∣∣∣< ∑j

m j = n

and so, you would have fewer than n vectors in this basis of eigenvectors which cannotoccur. Hence A is defective. Thus if an eigenvalue is defective, the matrix is defective andif no eigenvalue is defective, then the matrix is nondefective. ■

12.1.6 DiagonalizationFirst of all, here is what it means for two matrices to be similar.

Definition 12.1.20 Let A,B be two n×n matrices. Then they are similar if and only if thereexists an invertible matrix S such that

A = S−1BS

Proposition 12.1.21 Define for n×n matrices A∼ B if A is similar to B. Then

A∼ A,

If A∼ B then B∼ A

If A∼ B and B∼C then A∼C