Download Generalized Eigenvectors

5.9 Generalized Eigenvectors
In the previous section we looked at the case where each eigenvalue of a square matrix A
has as many linearly independent eigenvectors as its multiplicity. In that case we could
diagonalize A and use this to solve differential and difference equations where A is the
coefficient matrix.
Now we turn to situations where some eigenvalues do not have as many linearly
independent eigenvectors as their multiplicities. In such cases we shall use generalized
eigenvectors as a substitute for regular eigenvectors. They can be used to express A as
A = TJT-1 where J, called the Jordan canonical form of A, is "almost" diagonal. This
generalization of diagonalization can be used to solve differential and difference
equations in a fashion similar to the previous sections. Let's look at an example.
1 1
Example 1. Let A = -1 3. Find the eigenvalues and eigenvectors of A.
For the eigenvalues one has
1- 1
0 =  -1 3 -  = (1 - )(3 - ) + 1 = 2 - 4 + 4 = ( - 2)2
x
The only eigenvalue is  = 2 which is of multiplicity two. An eigenvector v =  y 
satisfies
 0  = (A - 2I)v = -1 1  x 
0
-1 1  y 
Both equations are - x + y = 0 or y = x so the eigenvectors are multiples of
1
v1 =  1 
So even though the eigenvalue  = 2 is of multiplicity two it has only one linearly
independent eigenvector.
In situations like this we turn to generalized eigenvectors as a substitute for eigenvectors.
In order to appreciate the definition of a generalized eigenvector, note that an eigenvector
v for an eigenvalue  satisfies (A - I)v = 0. If we replace A - I by (A - I)m we get the
definition of a generalized eigenvector.
5.9 - 1
Definition 1. Let A be a square matrix and  be an eigenvalue of A. A vector v is a
generalized eigenvector for  if
1.
v  0
2.
(A - I)mv = 0 for some positive integer m
The smallest positive integer m such that (A - I)mv = 0 is called the degree of the
generalized eigenvector.
Note that any eigenvector of A is a generalized eigenvector of degree one since it satisfies
(A - I)mv = 0 with m = 1. In fact an eigenvector v of A satisfies (A - I)mv for any
positive integer m since (A - I)mv = (A - I)m-1(A - I)v = (A - I)m-10 = 0. More
generally, one has the following.
Proposition 1. If v is a generalized eigenvector for the eigenvalue  of degree m then
(A - I)pv = 0 for every positive integer p that is larger than m.
Proof. If (A - I)mv = 0 then (A - I)m+kv = (A - I)k(A - I)mv = (A - I)k0 = 0. //
1 1
Example 2. Find the generalized eigenvectors of A = -1 3.
In Example 1 we saw that  = 2 was the only eigenvalue and the only eigenvectors were
1
multiples of v1 =  1 . These are the generalized eigenvectors of degree one.
The generalized eigenvectors of degree two are the solutions of (A - 2I)2v = 0. There are
two ways to find these. The first is to compute (A - 2I)2 and solve (A - 2I)2v = 0. One has
-1 1
-1 1 -1 1
0 0
A - 2I = -1 1 so (A - 2I)2 = -1 1 -1 1 = 0 0 = 0. So the equation (A - 2I)2v = 0 is
x
0v = 0. Every vector v =  y  satisfies this equation, so every non-zero vector v is a
1
generalized eigenvector of A. The ones that do not lie on the line through v1 =  1  are of
degree two.
The second way to find the generalized eigenvectors of degree two is to note that if v is a
generalized eigenvector of degree two then u = (A - 2I)v is a generalized eigenvector of
degree one since (A - 2I)u = (A - 2I)2v = 0. So u = cv1 for some number c and
(A - 2I)v = cv1. We can find the solutions to this equation by solving (A - 2I)v = v1 and
x
then multiplying v by c. If we let v =  y  then the equation (A - 2I)v = v1 is
-1 1  x  =  1 . Both equations are - x + y = 1, so y = x + y and
-1 1  y   1 
(1)
v =
x
y
x
=  x + 1 
0
1
0
=  1  + x  1  =  1  + xv1
5.9 - 2
for any x. The non-zero multiples of these vectors are the generalized eigenvectors of
degree 2. It is not hard to see that any vector, except the multiples of v1, can be written as
a multiple of the vectors v in (1).
The second method of computing the generalized eigenvectors of degree two is a special
case of the following more general result.
Proposition 2. v is a generalized eigenvector of degree m if and only if (A - I)kv is a
generalized eigenvector of degree m – k for k = 1, 2, …, m - 1. In particular, v is a
generalized eigenvector of degree m if and only if Av = v + w where w is a generalized
eigenvector of degree m – 1.
Proof. One has (A - I)mv = (A - I)m-k(A - I)kv. So if v is a generalized eigenvector of
degree m then (A - I)mv = 0. So (A - I)m-k(A - I)kv = 0 and (A - I)kv has degree at
most m - k. If it has degree p and p is less than m - k then (A - I)p(A - I)kv = 0 which
implies (A - I)p+kv = 0 which implies v has degree less than m. This proves the only if
part of the first assertion. The if part is proved in a similar fashion. The second assertion
follows from the first. //
The generalized eigenvectors can be used to create a substitute for the diagonalization of
the matrix. We start with a 22 matrix A with a single eigenvalue  that has only one
linearly independent eigenvector v1 as in Example 1. It turns out that A must have a
generalized eigenvector v of degree two. So (A - I)v is a regular eigenvector. So
(A - I)v = cv1. Let v2 = v/c. Then (A - I)v2 = v1 or Av2 = v1 + v2. Let
T = matrix whose columns are v1 and v2
 1
J = 0 
Since Av1 = v1 and Av2 = v1 + v2 the matrix AT has columns equal to v1 and v1 + v2.
Consider TJ. Since that the kth column of TJ is the linear combination of the columns of
T using the entries in the kth column of J as coefficients, TJ has columns equal to v1 and
v1 + v2. So AT = TJ or
(2)
 1
A = TJT-1 = T 0  T-1
 1
J = 0  is called the Jordan canonical form of A and we shall call formula (2) the
Jordanization of A. It is a substitute for the diagonalization of A when it comes to
various computations with A.
5.9 - 3
More generally, suppose we have an nn matrix A with a generalized eigenvector vn of
degree n for the eigenvalue . We can let
vn-1 = (A - I)vn
vn-2 = (A - I)vn-1
...
...
v2 = (A - I)v3
v1 = (A - I)v1
which has degree n - 1
which has degree n - 2
which has degree 2
which has degree 1, i.e. is a regular eigenvalues
Then
(3)
Avn = vn-1 + vn
Avn-1 = vn-2 + vn-1
...
...
Av3 = v2 + v3
Av2 = v1 + v2
Av1 = v1
Let
T = matrix whose columns are v1, v2, …, vn
 1 0
0  1
0 0 
(4)
J =

.
0.
0
0 0
0 0



 1
0 
0 0
0 0
0 0
matrix with ’s on the main diagonal
= 1's one the diagonal above the main
diagonal and 0's elsewhere
In view of (3) the matrix AT has columns equal to v1, v1 + v2, …, vn-1 + vn. Consider
TJ. Since that the kth column of TJ is the linear combination of the columns of T using
the entries in the kth column of J as coefficients, TJ has columns are also equal to
v1, v1 + v2, …, vn-1 + vn. So AT = TJ or
 1 0
0  1
0 0 
(5)
A = TJT-1 =


T .
0.
0

T

 1
0 
0 0
0 0
0 0
0 0
0 0
-1
Again J is the Jordan canonical form of A and (4) is the Jordanization of A.
5.9 - 4
Definition 2. A matrix of the form (4) is called a Jordan block matrix.
For a general non-diagonalizable matrix the Jordan canonical form looks like
J =






J1
0
0
.
.
0
0
.
.
0
0
0 0
J2 0
0 J3
0 0 0
0 0 0
0 0 0
0 0
0 0
0 Jr 0
0 0 r+1
0 0
0 0
0 0 0
0 0 0


0 0 
0 0


 1
0 
0 0
0 0
0 0
s-1
s
Each Jk is a Jordan block of the form (4) which is a mkmk matrix, so
Jk =




k 1 0
k 1
0 k
0
0
.
.
0
0
0 0
0 0



 1
0 
0 0
0 0
0 0
k
k
Each Jk is associated with a series of generalized eigenvectors vk,1, …, vk,mk where vk,1 is a
regular eigenvector and vk,j-1 = (A - kI)vk,j for j = 2, …, mk. The columns of T are the vk,j
which are all independent. Finding the vk,j requires a bit of work. We will discuss the
general case more after we look at some examples.
1 1
Example 3. Let A = -1 3. Find the Jordanization (2) of A.
In Example 1 we saw that  = 2 was the only eigenvalue and the only eigenvectors are
1
0
1
multiples of v1 =  1 . In (2) we saw that v2 =  1  + x  1  is the solution to
0
(A - I)v2 = v1 where x can be any number. If we take x = 0 then v2 =  1  and the
Jordanization of A is
-1
 1 1 = 1 0 2 1 1 0
-1 3
1 1 0 2 1 1
This can be used to compute the powers of a matrix A. As in section 5.2 one has
5.9 - 5
 1 n
An = TJnT-1 = T 0  T-1
 1 n
It turns out that 0  is quite simple. One has
2
2
 1 =  1  1 =  22 
0  
0 
 0   0 
3
2
2
2
3
 1 =  1  1 =  22   1 =  33 
 0   0 
0  
0 
0  0 
4
3
2
3
3
4
 1 =  1  1 =  33   1 =  44 
 0   0 
0  
0 
0  0 
and in general
n
n-1
n
 1 =  nn 
0  
0 
So
n nn-1
An = T  0 n  T-1
1 1
Example 4. Let A = -1 3. Find An.
1 1
1
From Example 3 one has -1 3 = 1
0 2 1 1 0-1
1 0 2 1 1 ,
so
n
-1
n
n-1
 1 1 = 1 0 2 n2n  1 0 = 2n 1 n/2   1 0
-1 3
1 1  0 2  1 1
1 n/2 + 1 -1 1
1
n/2
n/2
= 2n  - n/2 n/2 + 1
The change of variables method used in sections 5.3 and 5.4 to decouple systems of
difference and differential equations can be used when the coefficient matrix has
generalized eigenvectors. The matrix T in the Jordanization is used as a substitute for the
matrix T in the diagonalization. The new equations are not completely decoupled, but
one doesn’t involve the other. We solve that one first and then the other.
1 1
Example 5. Use the Jordanization of A = -1 3 to solve the difference equations
xn+1 =
xn +
yn + 3
x0 = 4
xn+1 = - xn + 3yn - 2
y0 = 5
5.9 - 6
The difference equations can be written as
 xn+1  =  1
 yn+1 
-1
1   xn 
3   yn 
3
+  - 2 
 x0  =  4 
 y0 
5
4
x
3
or un+1 = Aun + b with u0 =  5  where un =  ynn  and b =  - 2 . We make the change of
1 0  pn 
1 0
variables un = Tvn = 1 1  q  where T = 1 1 is the matrix whose columns are the
n
eigenvectors and generalized eigenvectors of A. Then the equations become
4
v0 = T-1 5 
vn+1 = Jvn + c
2 1
where J = 0 2 is the canonical form of A and
1 0
3
3
c = T-1b = -1 1  - 2  =  - 5 
4
v0 = T-1 5  =
 1 0  4  =  4 
-1 1  5 
1
So vn+1 = Jvn + c becomes
 pn+1  = 2 1  pn  +  3 
 qn+1 
0 2  qn 
-5
 pn  =  4 
 qn 
1
or
pn+1 =
2pn + qn + 3
p0 = 4
qn+1 =
2qn - 5
q0 = 1
This new pair of equations is not completely decoupled, but the equation for qn does not
involve pn. We can solve the equation for qn first and substitute the solution into the
equation for pn and then solve that.
For the equation for qn the solution to the homogeneous equation qn+1 = 2qn is qn = C 2n
where C is any constant. A particular solution has the form qn = g. Plugging into
qn+1 = 2qn - 5 we get g = 2g - 5. So g = 5 and the general solution to qn+1 = 2qn - 5 is
qn = 5 + C 2n. We use the initial condition q0 = 1 to find C. So 1 = 5 + C. So C = - 4
and qn = 5 - 42n.
We substitute this into the equation for pn which becomes
(6)
pn+1 =
2pn + 8 - 42n
p0 = 4
5.9 - 7
The solution to the homogeneous equation pn+1 = 2pn is pn = C 2n where C is any
constant. Since 2n is a solution to the homogeneous equation, a particular solution to (6)
has the form pn = g + hn 2n. Plugging into (6) we get
g + h(n+1) 2n+1 = 2(g + h 2n) + 8 - 42n
Equating coefficients we have g = 2g + 8 and 2h = - 4. So g = - 8 and h = - 2 and a
particular solution is pn = - 8 - 2n 2n. Adding the solution to the homogeneous equation
gives pn = - 8 - 2n 2n + C 2n. Using the initial condition gives 4 = - 8 + C. So C = 12 and
pn = - 8 - 2n 2n + 122n. So
p
-- 8 - 2n 2 + 122 
vn =  qnn  = 

5 - 42n
n
n
and
n
n
n
n
 xn  = T pn  = 1 0  -- 8 - 2n 2 +n122  =  - 8 - 2n 2 n+ 122n 

 - 3 - 2n 2 + 82 
 yn 
 qn 
1 1 
5 - 42
 4 - 3 - 2
Example 6. Find the eigenvalues and generalized eigenvectors of A =  8 - 6 - 4. Also
- 4 3 2 
find the Jordanization of A.
For the eigenvalues one has
-2 
4 -  - 3
- 
0 =  8 -6- -4  =  8
-4
- 4
3 2 - 
 1
= - 0
- 4
0
-
3
1
= 2 3
2 
6 - 
1 
- 2
2 - 
0
-6-
3
 1
= 2  0
- 4
- 
-4 
2 - 
0
1
3
 1
= - 8
- 4
1 
2 
2 - 
1 
-4 
2 - 
0
-6-
3
 1
= 2  0
- 4
0
1
3
0 
2 
6 - 
= - 3
x
The only eigenvalue is  = 0 which is of multiplicity three. An eigenvector u =  y 
z
satisfies
0
 4 - 3 - 2  x 
0
  = Av =  8 - 6 - 4  y 
0
- 4 3 2   z 
or
4x - 3y - 2z = 0
8x - 6y - 4z = 0
- 4x + 3y + 2z = 0
5.9 - 8
Note that equation 2 is twice equation 1 and equation 3 is the negative of equation 1. So
if we solve equation 1, we solve both equations. The first equation can be written as
x
 3/4 
 1/2 
x = 34 y + 12 z. So u =  y  = y  1  + z  0 . So the eigenvectors are linear
z
0 
1 
combinations of
u1
3
= 4
0
1
u2 =  0 
2
So the eigenvalues of A corresponding to  form the two dimensional plane through u1
and u2.
x
To find a generalized eigenvectors w =  y  of degree 2, we use the fact that Aw is a
z
regular eigenvector. So Aw = au1 + bu2 for some numbers a and b. Writing this out as
equations one has
4x - 3y - 2z = 3a + b
8x - 6y - 4z = 4a
- 4x + 3y + 2z = 2b
Adding equation 1 to equation 3 and subtracting 2 times equation 1 from equation 2 gives
4x - 3y - 2z = 3a + b
0 = - 2a - 2b
0 = 3a + 3b
The last two equations each say b = - a. Then the first equation becomes
4x - 3y - 2z = 2a
x
 1/2   3/4 
 1/2 
y
1
1
So x = a + y + z. So w =   = a   y   + z  0 . If a = 0 then w is just a
z
0  0 
1 
regular eigenvector. If a  0 then w is a generalized eigenvector of degree two. So the
x
generalized eigenvectors of degree two are all vectors  y  that do not lie on the plane of
z
eigenvectors through u1 and u2. Since this is all of three dimensional space there are no
generalized eigenvectors of degree higher than two.
1
2
3
4
1
2
Since zero is the only eigenvalue and it has a two dimensional plane of eigenvectors and
the vectors not in this plane are generalized eigenvectors of degree two, the Jordanization
of A will have the form
5.9 - 9
 1 0
0 1 0
A = TJT-1 = T 0  0 T-1 = T 0 0 0 T-1
0 0 0
0 0 
and the columns of T will be v11, v12 and v1 where v12 is any generalized eigenvector of
degree two and v11 = Av12 and v2 are regular eigenvectors. For v12 we can choose any
0
-2
vector not on the plane of eigenvectors. Let’s choose v12 =  0 . The v11 = Av12 =  - 4 .
1
2 
Then for v2 we can choose any eigenvector that is not a multiple of v11. Let’s choose
1
-2 0 1 
v2 = u2 =  0 . Then T is the matrix with columns v11, v12 and v2, i.e T = - 4 0 0  So
2
 2 1 2
The Jordanization of A is
A = TJT
-1
-1
 - 2 0 1  0 1 0  - 2 0 1 
=  - 4 0 0  0 0 0  - 4 0 0 
 2 1 2  0 0 0  2 1 2 
Here are some properties of generalized eigenvectors that are used to obtain the Jordan
canonical form of a general non-diagonalizable matrix.
Definition 2. Let A be a square matrix and  be an eigenvalue of A. Let
N,m = {v: (A - I)mv = 0} be the set of generalized eigenvectors for  of degree at most m
along with the zero vector. N,m is called the generalized eigenspace of degree m for .
The generalized eigenspaces are subspaces since they are the null spaces of the matrices
(A - I)m.
Proposition 3. Let A be a square matrix and  be an eigenvalue of A and N,m be the
generalized eigenspace of degree m for .
(a) N,1  N,2   N,m  N,m+1  
(b)
(c)
If N,m = N,m+1 for some m then N,m = N,m+k for all positive integers k, i.e.
N,m = N,p for all positive integers p larger than m.
There exists r such that N,1   N,2   N,r-1  N,r and
N,r = N,r+1 = N,r+2 =  = N,m =  for m > r.
Proof. (a) follows from Proposition 1. To show (b) we shall show that N,m = N,m+1
implies N,m+1 = N,m+2. Applying this over and over again will show N,m = N,m+k for all
positive integers k. Note that N,m = N,m+1 is equivalent to
(A - I)m+1v = 0  (A - I)mv = 0. Suppose (A - I)m+2v = 0. We write
(A - I)m+2v = (A - I)m+1(A - I)v. So (A - I)m+1(A - I)v = 0. So (A - I)m(A - I)v = 0.
So (A - I)m+1v = 0. Thus (A - I)m+2v = 0  (A - I)m+1v = 0. So N,m+1 = N,m+2 which
proves (b). (c) follows from (b). //
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