6.3 Eigenvalues And Eigenvectors Of Linear Transformations
We begin with the following fundamental definition.
Definition 6.3.1 Let L ∈ℒ
where V is a vector space of dimension n with field of scalars F. An
eigen-pair consists of a scalar λ ∈ F called an eigenvalue and a NON-ZERO v ∈ V such
Do eigen-pairs exist? Recall that from Theorem 6.1.10 the minimum polynomial can be factored in a
unique way as
where each ϕi
is irreducible and monic. Then the following theorem is obtained.
Theorem 6.3.2 Let L ∈ℒ
and let its minimum polynomial p
have a root μ in the field
of scalars. Then μ is an eigenvalue of L.
Proof: Since p
has a root, we know
where the degree of
is less than the
. Therefore, there is a vector
such that q
u ≡ v≠
is not really the
minimum polynomial. Then
= 0 and so μ
is indeed an eigenvalue.
Theorem 6.3.3 Suppose the minimum polynomial of L ∈ℒ
factors completely into linear
factors (splits) so that
Then the μi are distinct eigenvalues and corresponding to each of these eigenvalues, there is an
eigenvector wi≠0 such that Lwi = μiwi. Also, there are no other eigenvalues than these μi.
and if Li is the restriction of L to ker
ki, then Li has exactly one eigenvalue and it is
Proof: By Theorem 6.3.2, each μi is an eigenvalue and we can let wi be a corresponding eigenvector.
By Theorem 6.1.10,
Also by this theorem, the minimum polynomial of Li is
and so it has an eigenvalue μi
. Could Li
have any other eigenvalue ν≠μi
? To save notation, denote by m
the exponent ki
and by μ
the eigenvalue μi
Also let w
denote an eigenvalue of Li
with respect to ν
. Then since minimum polynomial for Li
which is impossible because w≠
0. Thus there can be no other eigenvalue for Li
. Consider the claim about L
having no other eigenvalues than the μi
. Say μ
is another eigenvalue with eigenvector w
. Then let
. Then not every zi
= 0 and
Since this is a direct sum and each ker
is invariant with respect to L
, we must have
each Lizi − μzi
This is impossible unless μ
equals some μi
because not every zi
Example 6.3.4 The minimum polynomial for the matrix
is λ2 − 3λ + 2. This factors as
and so the eigenvalues are
2. Find the eigen-pairs. Then
determine the matrix with respect to a basis of these eigenvectors if possible.
First consider the eigenvalue 2. There exists a nonzero vector v such that
= 0. This follows
from the above theory. However, it is best to just find it directly rather than try to get it by using the proof
of the above theorem. The augmented matrix to consider is then
Row reducing this yields
Thus the solution is any vector of the form
Now consider the eigenvalue 1. This time you row reduce
which yields for the row reduced echelon form
Thus an eigenvector is of the form
Consider a basis for ℝn of the form
You might want to consider Problem 9 on Page 298 at this point. This problem shows that the matrix with
respect to this basis is diagonal.
When the matrix of a linear transformation can be chosen to be a diagonal matrix, the transformation
is said to be nondefective. Also, note that the term applies to the matrix of a linear transformation
and so I will specialize to the consideration of matrices in what follows. As shown above, this
is equivalent to saying that any matrix of the linear transformation is similar to one which
is diagonal. That is, the matrix of a linear transformation, or more generally just a square
matrix A has the property that there exists S such that S−1AS = D where D is a diagonal
Here is a definition which also introduces one of the most horrible adjectives in all of mathematics.
Definition 6.3.5 Let A be an n×n matrix. Then A is diagonalizable if there exists an invertible matrix
S such that
where D is a diagonal matrix. This means D has a zero as every entry except for the main diagonal. More
precisely, Dij = 0 unless i = j. Such matrices look like the following.
where ∗ might not be zero.
The most important theorem about
is the following major result. First here is a simple observation.
Observation 6.3.6 Let S =
where S is n×n. Then here is the result of multiplying on
the right by a diagonal matrix.
This follows from the way we multiply matrices. The ith entry of the jth column of the product on the left is
of the form siλj. Thus the jth column of the matrix on the left is just λjsj.
Theorem 6.3.7 An n × n matrix is diagonalizable if and only if Fn has a basis of eigenvectors of A.
Furthermore, you can take the matrix S described above, to be given as
where here the sk are the eigenvectors in the basis for Fn. If A is diagonalizable, the eigenvalues of A are
the diagonal entries of the diagonal matrix.
Proof: To say that A is diagonalizable, is to say that for some S,
the λi being elements of F. This is to say that for S =
being the kth
which is equivalent, from the way we multiply matrices, that
which is equivalent to saying that the columns of S are eigenvectors and the diagonal matrix has the
eigenvectors down the main diagonal. Since S−1 is invertible, these eigenvectors are a basis. Similarly, if
there is a basis of eigenvectors, one can take them as the columns of S and reverse the above steps, finally
concluding that A is diagonalizable. ■
Corollary 6.3.8 Let A be an n × n matrix with minimum polynomial
Then A is diagonalizable if and only if each ki = 1.
Proof: Suppose first that it is diagonalizable and that a basis of eigenvectors is
. Since n ≥ p,
there may be some repeats here, a μi
going with more than
. Say ki >
Thus this is a monic
polynomial which has smaller degree than p
If you have v ∈ Fn,
since this is a basis, there
are scalars ci
such that v
is arbitrary, this shows that
= 0 contrary to the definition of the minimum polynomial being
. Thus each
Conversely, if each ki = 1, then
and you simply let βi be a basis for ker
which consists entirely of eigenvectors by definition of
what you mean by ker
Then a basis of eigenvectors consists of
and so the
is diagonalizable. ■
Example 6.3.9 The minimum polynomial for the matrix
is λ3 − 5λ2 + 8λ − 4. This factors as
and so the eigenvalues are
2. Find the
eigen-pairs. Then determine the matrix with respect to a basis of these eigenvectors if possible. If
it is not possible to find a basis of eigenvectors, find a block diagonal matrix similar to the
First find the eigenvectors for 2. You need to row reduce
Thus the eigenvectors which go with 2 are
The eigenvectors which go with 1 are
By Theorem 6.3.3, there are no other eigenvectors than those which correspond to eigenvalues
1,2. Thus there is no basis of eigenvectors because the span of the eigenvectors has dimension
However, we can consider
The second of these is just span
What is the first? We find it by row reducing the
following matrix which is the square of A −
augmented with a column of zeros.
Row reducing this yields
which says that solutions are of the form
This is the nonzero vectors of
What is the matrix of the restriction of A to this subspace?
and so some computations yield
Indeed this works
Then the matrix associated with the other eigenvector is just 1. Hence the matrix with respect to the above
ordered basis is
So what are some convenient computations which will allow you to find M easily? Take the transpose of
both sides of 6.2. Then you would have
and so MT =
The eigenvalue problem is one of the hardest problems in algebra because of our inability to exactly
solve polynomial equations. Therefore, estimating the eigenvalues becomes very significant. In the case of
the complex field of scalars, there is a very elementary result due to Gerschgorin. It can at least give an
upper bound for the size of the eigenvalues.
Theorem 6.3.10 Let A be an n × n matrix. Consider the n Gerschgorin discs defined as
Then every eigenvalue is contained in some Gerschgorin disc.
This theorem says to add up the absolute values of the entries of the ith row which are off the main
diagonal and form the disc centered at aii having this radius. The union of these discs contains
Proof: Suppose Ax = λx where x≠0. Then for A =
Now dividing by
, it follows
is contained in the kth
Gerschgorin disc. ■
In these examples given above, it was possible to factor the minimum polynomial and explicitly
determine eigenvalues and eigenvectors and obtain information about whether the matrix was
diagonalizable by explicit computations. Well, what if you can’t factor the minimum polynomial? What
then? This is the typical situation, not what was presented in the above examples. Just write down a 3 × 3
matrix and see if you can find the eigenvalues explicitly using algebra. Is there a way to determine whether
a given matrix is diagonalizable in the case that the minimum polynomial factors although you might have
trouble finding the factors? Amazingly, the answer is yes. One can answer this question completely using
only methods from algebra.