In this book, we are tacitly assuming that the field of scalars is ℂ so that the characteristic polynomial can always be factored.
Definition A.7.4 By the fundamental theorem of algebra, there exists a factorization of the characteristic equation in the form

where r_{i} is some integer no smaller than 1. Thus the eigenvalues are λ_{1},λ_{2},
Example A.7.5 Consider the matrix
 (1.30) 
What is the algebraic multiplicity of the eigenvalue λ = 1?
In this case the characteristic equation is

or equivalently,

Therefore, λ is of algebraic multiplicity 3.
Example A.7.7 Find the geometric multiplicity of λ = 1 for the matrix in 1.30.
We need to solve

The solutions are of the form

It follows the geometric multiplicity of λ = 1 is 1.
Definition A.7.8 An n×n matrix is called defective if the geometric multiplicity is not equal to the algebraic multiplicity for some eigenvalue. Sometimes such an eigenvalue for which the geometric multiplicity is not equal to the algebraic multiplicity is called a defective eigenvalue. If the geometric multiplicity for an eigenvalue equals the algebraic multiplicity, the eigenvalue is sometimes referred to as nondefective.
Here is another more interesting example of a defective matrix.
Example A.7.9 Let

Find the eigenvectors and eigenvalues.
In this case the eigenvalues are 3,6,6 where we have listed 6 twice because it is a zero of algebraic multiplicity two, the characteristic equation being

Then a computation shows that the geometric multiplicity of 6 is 1.
Consider the eigenvectors for λ = 6. This requires you to solve

and the augmented matrix for this system of equations is

Then from row operations,

and so the eigenvectors for λ = 6 are of the form

where t ∈ F.
Note that in this example the eigenspace for the eigenvalue λ = 6 is of dimension 1 because there is only one parameter. However, this eigenvalue is of multiplicity two as a root to the characteristic equation. Thus this eigenvalue is a defective eigenvalue. However, the eigenvalue 3 is nondefective. The matrix is defective because it has a defective eigenvalue.
The word, defective, seems to suggest there is something wrong with the matrix. This is in fact the case. Defective matrices are a lot of trouble in applications and we wish they never occurred. However, they do occur as the above example shows. The reason these matrices are so horrible to work with is that it is impossible to obtain a basis of eigenvectors.
In terms of algebra, this lack of a basis of eigenvectors says that it is impossible to obtain a diagonal matrix which is similar to the given matrix. See below for this definition.
Although there may be repeated roots to the characteristic equation, it is not known whether the matrix is defective. However, there is an important theorem which holds when considering eigenvectors which correspond to distinct eigenvalues.
Theorem A.7.10 Suppose Mv_{i} = λ_{i}v_{i},i = 1,
Proof. Suppose the claim of the lemma is not true. Then there exists a subset of this set of vectors

such that
 (1.31) 
where each c_{j}≠0. Say Mw_{j} = μ_{j}w_{j} where

the μ_{j} being distinct eigenvalues of M. Out of all such subsets, let this one be such that r is as small as possible. Then necessarily, r > 1 because otherwise, c_{1}w_{1} = 0 which would imply w_{1} = 0, which is not allowed for eigenvectors.
Now apply M to both sides of 1.31.
 (1.32) 
Next pick μ_{k}≠0 and multiply both sides of 1.31 by μ_{k}. Such a μ_{k} exists because r > 1. Thus
 (1.33) 
Subtract the sum in 1.33 from the sum in 1.32 to obtain

Now one of the constants c_{j}