We’ve investigated two by two matrices with
only a single eigenvalue, which produces only
a single eigenvector. In this video, we introduce
the problem of generalizing the method, and
define complete and defective eigenvalues.
Remember the two-by-two case. We have a single
eigenvalue, lambda. It only gives rise to
a single eigenvector, v. We can use that to
produce a single solution, v times e to the
lambda t.
For the other solution, we needed to find
a special vector w. w satisfies an eigenvector-like
condition with the eigenvector v; in particular,
(A minus lambda I)v = 0
(A minus lambda I)w = v
We then created a new solution, (vt plus w)e
to the lambda t.
Now, consider an arbitrary square matrix A.
The characteristic polynomial of A factors
completely over the complex numbers:
P of lambda equals (lambda – lambda 1) to
the r_1 times (lambda – lambda 2) to the
r_2 and so on.
These lambda_i are the eigenvalues, and these
multiplicities of the roots, are the multiplicities
of the eigenvalues.
From each eigenvalue, we need to get 
as many solutions as its multiplicity. So
if for example, we have a (lambda minus two)
raised to the three, then two is an eigenvalue,
and we need to get three solutions from it.
We actually get as many solutions as there
are linearly independent eigenvectors associated
with the eigenvalue. That could be three,
in which case, good, we’re done with the
eigenvalue. But it could also be two or one.
Not zero, every eigenvalue has eigenvectors
associated with it.
We call an eigenvector of multiplicity k complete
if it has k linearly independent eigenvectors.
Otherwise, it is defective, and the 
defect is k minus the number of linearly independent
eigenvectors it does have.
We have worked with the two-by-two case already.
In that case, if an eigenvalue is defective,
its defect is automatically one. We are now
prepared to generalize that method.
