Transcript Eigenvalues and Eigenvectors

Section 4.1 Eigenvalues and Eigenvectors
Recall that a matrix transformation is, a function f from Rn to Rn determined by an
n  n matrix A so that f(p) = Ap, for p in Rn.
The length of input vector p need not be the same as the length of the image
vector Ap; that is ||p|| need not equal ||Ap||.
Since a vector p in Rn has two basic properties, length and direction, it is reasonable
to investigate whether p and Ap are ever parallel for a given matrix A. (Recall that
parallel means that one vector is a scalar multiple of the other; Ap = (scalar) p .)
We note that A0 = 0, hence we need consider only nonzero vectors p.
The vectors p such that Ap and p are parallel arise in a surprising number of
places, including Markov problems, population model, differential equations,
vibrations, aerodynamics, elasticity, nuclear physics, mechanics, chemical
engineering, biology, digital transmission of images, etc.
Here we consider only square matrices. We begin with 2 × 2 and 3 × 3 matrices,
but our techniques also apply to larger square matrices.
Example:
Show that Ap is parallel to p and that Aq is parallel to q.
Example:
Example:
Strategy: Compute Ap and set it equal to scalar λ times p. Use equality of vectors
to determine λ and the components of vector p.
We have

x = λx & -y = λy
Setting corresponding entries equal we have two cases to investigate:
Case1: x = λx
So for x not 0 we have λ =1; then for –y = λy we have –y = y so that y = 0.
Case 2: -y = λy
So for y not 0 we have λ = -1; then for x = λx we have x = -x, so that x = 0.
For matrix
we have the set of vectors p so that Ap =λp,
are all vectors in R2 of the form
Note that only particular values of scalar λ can be chosen.
Definition Let A be an n  n matrix. The scalar  is called an eigenvalue of matrix A
if there exists an n  1 vector x, x  0, such that
Ax = x.
Every nonzero vector x satisfying (1) is called an eigenvector of A associated with
eigenvalue .
Equation Ax = x is commonly called the eigen equation and its solution is a goal of
this chapter.
Sometimes we will use the term eigenpair to mean an eigenvalue and an associated
eigenvector. This emphasizes that for a given matrix A a solution of (1) requires both a
scalar, an eigenvalue , and a nonzero vector, an eigenvector x.
In the eigen equation Ax = x there is no restriction on the entries of matrix A, the
type of scalar , or the entries of vector x, other than they cannot all be zero. It is
possible that eigenvalues and eigenvectors of a matrix with real entries can involve
complex values.
Example: For matrix
we found eigenpairs
Note that there are many eigenvectors associated with a particular eigenvalue.
Example: For matrix
is an eigenpair and that
Example: For matrix
show that
is another eigenpair.
show that
is an eigenpair
Properties:
If x is an eigenvector corresponding to eigenvalue  of A, then so is kx for any
scalar k  0. (A nonzero scalar multiple of an eigenvector is another
eigenvector corresponding to the same eigenvalue.)
Proof: If (,p) is an eigenpair of A then Ap = p. We know that p  0 and k  0 so it
must be that kp  0. To show that kp is an vector we do the following:
A(kp) = k(Ap) = k(p) = (kp),
This shows that (,kp) is an eigenpair of A.
If p and q are eigenvectors corresponding to eigenvalue  of A, then so is p + q
(assuming p + q  0).
Proof: If (,p) is an eigenpair of A then Ap = p.
If (,q) is an eigenpair of A then Aq = q.
To show that p + q is an an eigenpair we do the following:
A(p + q) = Ap + Aq = p + q = (p + q)
This shows that (,p + q) is an eigenpair of A.
Properties: (continued)
If (,p) is an eigenpair of A, then for any positive integer r, (r,p) is an eigen pair of Ar.
Proof: Since (,p) is an eigenpair of A then Ap = p. Thus we have
A2p = A(Ap) = A(p) = (Ap) = (p) = 2p,
A3p = A(A2p) = A(2p) = 2(Ap) = 2(p) = 3p,
and in general
Arp = A(Ar-1p) = A(r-1p) = r-1(Ap) = r-1(p) = rp
If (,p) and (,q) are eigenpairs of A with   , then p and q are linearly independent.
Proof: Here we use material from Section 2.4, namely that if vectors p and q are not
scalar multiples of one another, then they are linearly independent.
Since (,p) is an eigenpair of A then Ap = p. Similarly since (,q) is an eigenpair of A
then Aq = q. Let s and t be scalars so that tp + sq = 0, or equivalently tp = -sq.
Multiplying this linear combination by A we get
A( tp + sq) = t(Ap) + s(Aq) = t(p) + s(q) = (tp) + (sq) = 0 .
Substituting for tp we get (-sq) + (sq) = 0 or equivalently that s( - )q = 0. Since
q  0 and   , it must be that s = 0, but then tp = -0q = 0. It follows that since p  0
then t = 0. Hence the only way tp + sq = 0 is when t = s = 0, so p and q are linearly
independent.
(This property implies that eigenvector corresponding to distinct eigenvalues are
linearly independent.)
Let A be an n × n matrix with eigenpair (, p) . By definition we know that an
eigenvector cannot be the zero vector.
Next let S be the set of all eigenvectors of A corresponding to eigenvalue  together
with the zero vector.
Show that S is a subspace of Rn. (S is called the eigenspace corresponding to
eigenvalue .)
Application: Recall Markov processes from Section 1.3. The basic idea is that we
have a matrix transformation f(x) = Ax, where A is a square probability matrix; that is,
each entry of A is in the interval [0, 1] and the sum of the entries in each column is 1.
(Matrix A is called the transition matrix.)
In a Markov process the matrix transformation is used to form successive compositions
as follows:
If the sequence of vectors s0, s1, s2, ... converges to a limiting vector p, then we call p
the steady state of the Markov process. If the process has a steady state p, then
f(p) = p
if an only if
Ap = p.
That is, the steady state is an eigenvector corresponding to an eigenvalue  = 1 of the
transition matrix A.
Example:
Let be the transition matrix of a Markov process be
Show that the Markov process defined by A has a steady state and determine a
probability vector that represents the steady state distribution.
We seek a vector
we get
so that Ap = p. Using matrix algebra on the equation Ap = p
The solution to this homogeneous linear system gives
so we have eigenvector
, x  0, corresponding to eigenvalue  = 1.
We can choose a nonzero value for x so that vector p is a
probability vector. (That is, each entry of p is in the interval
[0, 1] and the sum of the entries is 1.) The value x = 3/8
works and so the steady state distribution is given by
3 
8 
p 
5 
 8 
Computing Eigen Information for Small Matrices
The eigen equation can be rearranged as follows:
Ax = x  Ax = Inx  Ax - Inx = 0  (A - In)x = 0
(1)
The matrix equation (A - In)x = 0 is a homogeneous linear system with coefficient
matrix A - In . Since an eigenvector x cannot be the zero vector, this means we seek a
nontrivial solution to the linear system (A - In)x = 0 . Thus ns(A - In)  0 or
equivalently rref(A - In) must contain a zero row. It follows that matrix A - In must be
singular, so from Chapter 3,
det(A - In) = 0.
(or det(In - A) = 0 )
(2)
Equation (2) is called the characteristic equation of matrix A and solving it for  gives
us the eigenvalues of A. Because the determinant is a linear combination of particular
products of entries of the matrix, the characteristic equation is really a polynomial in 
equation of degree n. We call
c() = det(A - In)
(3)
the characteristic polynomial of matrix A. The eigenvalues are the solutions of (2) or
equivalently the roots of the characteristic polynomial (3). Once we have the n
eigenvalues of A, 1, 2, ..., n, the corresponding eigenvectors are nontrivial solutions
of the homogeneous linear systems
(A - iIn)x = 0
for i = 1,2, ..., n.
(4)
We summarize the computational approach for determining eigenpairs (, x) as a
two-step procedure:
Example: Find eigenpairs of
Step I. Find the eigenvalues.
The eigenvalues are  1  3 and  2  2.
Step II. To find corresponding eigenvectors we solve
(A - iIn)x = 0
for i = 1,2
Given that  is an eigenvalue of A, then we know that matrix A - In is singular and
hence rref(A - In) will have at least one zero row.
A homogeneous linear system whose coefficient matrix has rref with at least one zero
row will have a solution set with at least one free variable. The free variables can be
chosen to have any value as long as the resulting solution is not the zero vector.
In the example for eigenvalue 1 = 3, the general solution was
The free variable in this case could be any nonzero value. We chose x2 = 2 to avoid
fractions, but this is not required. If we chose x2 = 1/7, then
is a valid eigenvector.
Since there will be at least one free variable when determining an eigenvector to
correspond to an eigenvalue, there are infinitely many ways to express the
entries of an eigenvector.
Property:
If x is an eigenvector corresponding to eigenvalue  of A, then so is kx for any
scalar k  0
Example:
Find the eigenvalues and corresponding eigenvectors of
The characteristic polynomial is
Its factors are
Corresponding eigenvectors are
So the eigenvalues are 1, 2, and 3.