Math 2280-001
Fri Mar 27
5.5 Linear systems x#= A x for which A is not diagonalizable.
We will spend most of the lecture finishing Wednesday's notes, about forced oscillations in mass-spring
systems, section 5.4. Then we'll begin section 5.5 in today's notes. It addresses the problem of how to
solve linear systems x#= A x when the matrix A is not diagonalizable. We will continue this discussion on
Monday next week.
preview/review: connecting the eigenvalue-eigenvector method for first order linear systems of
differential equations to change of basis ideas from Math 2270. Consider the first order system of
differential equations
x#= A x .
Case 1 If the matrix A is diagonalizable, then let S be a matrix whose columns are eigenvectors of A,
where these eigenvectors are a basis for =n or Cn . In the completely representative case of A2 # 2 we can
write S in column form,
S = v1 v2
with
A v1 = l 1 v1 , A v2 = l 2 v2 .
Then we have the identity
A S = A v1 v2 = l 1 v1 l 2 v2 = v1 v2
l1
0
0
l2
=AL
AS=SL.
where L is the diagonal matrix containing the eigenvalue of vj in the jth diagonal entry. This identity can be
solved either for A or L:
SK1 A S = L, A = S L SK1 .
This is the algebraic reason we call such matrices A "diagonalizable"- they are similar to diagonal matrices.
Now, return to the DE
x#= A x.
Consider the transformation
x t d S u t , equivalently SK1 x t = u t .
Substitute this into the system:
x#= A x
5 S u #= A S u
5 S u #= A S u.
(There is a universal product rule for any sort of multiplication that distributes over addition. We'll discuss
it Monday. In this case S is a constant matrix so the product rule reduces to the "constant multiple" rule for
differentiation above.)
5 u #= SK1 A S u
5 u #= L u
u1 #
5
=
u2 #
l1
0
0
l2
u1 t
l1u1
=
u2 t
l2u2
.
This diagonal system can be solved via Chapter 1:
l1 t
u1 #= l 1 u1 0 u1 t = c1 e
l2 t
u2 #= l 2 u2 0 u2 t = c2 e
.
Thus
u1 t
u2 t
l1 t
c1 e
=
.
l2 t
c2 e
Since x t d S u t ,
x t =
x1 t
x2 t
u1 t
= v1 v2
u2 t
l1 t
= v1 v2
c1 e
l2 t
l t
l2 t
= c1 e 1 v1 C c2 e
v2 .
c2 e
This is what we've been doing since the start of Chapter 5, but this time it's completely motivated, i.e. with
no "let's guess solutions of the form el t v" pulled out of thin air! This explanation is why that ad-hoc
method turned out to work.
Case 2 A is not diagonalizable. [I may ask you to read this on your own, and skip right to the exercises at
the end.] One learns in linear algebra that if An # n is not diagonalizable then instead of being similar to a
diagonal matrix, it is similar to a Jordan matrix, which is "almost" as good.
http://en.wikipedia.org/wiki/Jordan_normal_form
Here are some facts: For An # n Let the characteristic polynomial p l = det A K l I factor as
p l = K1
n
l K l1
k
1
k
l K l2
2
... l K l m
k
m
with m % n distinct eigenvalues l j , and k1 C k2 C...C km = n .
(1) Each eigenspace
El = nullspace A K l j I
j
has dimension
1 % dim El
j
% kj .
This dimension is called the geometric multiplicity of the eigenvalue l j , whereas the power kj is called the
algebraic multiplicity of l j .
(2) The matrix A is diagonalizable if and only if each dim El
j
= kj . In this case one can construct a
basis of =n (or Cn ) by amalgamating the various eigenspace bases. Algebraically this is expressed by the
identity
AS=SL
where the columns of S comprise a basis of =n (or Cn ) made out of eigenvectors of S, and L is the
diagonal matrix that has the corresponding eigenvalues in its columns. Each l j appears exactly kj times in
the diagonal matrix L, so the matrix L is unique up to how its columns (and S's columns are ordered. (In
this case the discussion at the start of the notes applies, for solving x#= A x.)
(3) Any eigenspace for which dim El
j
! kj is called defective. If A has any defective eigenspaces then
it is not diagonalizable. However, the larger generalized eigenspace
k
A K ljI
El 4 Gl d nullspace
j
j
j
does always have dimension kj . If bases for each Gl are amalgamated they will form a basis for =n or
j
Cn .
It is possible to choose the bases for each Gl so that when they are amalgamated into the columns of
j
a matrix S, the identity
AS=SJ
S A S = J; A = S J SK1
K1
holds, where J is the Jordan normal form of A. The structure of J is as follows: It's an upper triangular
matrix, and the diagonal entries are the eigenvalues of A. The eigenvalue l j appears exactly kj times on the
diagonal of J. If l j is defective, then there will be one or more square "Jordan" blocks centered on the
diagonal, having l j 's along the diagonal, ones along the super-diagonal, and zeroes everywhere else:
lj 1
0 lj
lj 1
,
0
0 l j 1 , etc.
0
0 lj
Off of the diagonal and super-diagonal all entries in J are zero. Jordan forms for a given matrix are
unique, up to reordering the blocks. For example, here is a possible Jordan normal form for an 8 # 8
matrix with characteristic polynomial
A K l I = l K l1
4
l K l2
2
l K l3
2
Exercise 1) Up to re-ordering the blocks, what are the other possible Jordan forms for an 8 # 8 matrix
having the same characteristic polynomial?
2×2 non-diagonalizable matrices: Because eigenvectors with different eigenvalues are always linearly
independent, the only way A2 # 2 can fail to be diagonalizable is if
A K l I = l K l1
with l 1 real, and dim El
2
= 1. In this case the Jordan form for A must be
1
l1 1
0 l1
.
And,
AS=SJ
reads
A S = A v1 v2 = v1 v2
l1
1
0
l1
= S J.
In other words,
Av1 = l 1 v1
A v2 = v1 C l 1 v2 .
Now repeat the transformation process that we did for the diagonalizable case...to solve the DE:
x#= A x.
Consider the transformation
x t d S u t , equivalently SK1 x t = u t .
Substitute this into the system:
x#= A x
5 S u #= A S u
5 S u #= A S u.
5 u #= SK1 A S u
5 u #= J u
u #
1
5 u # =
2
l1
1
0
l1
This system can be back-solved via Chapter 1:
l1u C u
u t
1
u t
2
=
1
l1u
2
2
.
l1 t
u2 #= l 1 u2 0 u2 t = c2 e
u1 #= l 1 u1 C u2
Kl t
1
e
l t
1
u1 #Kl 1 u1 = u2 = c2 e
Kl t l t
1
1
u1 #Kl 1 u1 = c2 e
Kl t
1
e
e
= c2
u1 = c2 t C c1
l t
1
u1 = c2 t C c1 e
.
Thus
l1 t
u1 t
=
u2 t
c2t C c1 e
.
l1 t
c2 e
Since x t d S u t ,
x t =
x1 t
x2 t
l1 t
u1 t
= v1 v2
c2t C c1 e
= v1 v2
u2 t
l t
l1 t
c2 e
l1 t
= c2t C c1 e 1 v1 C c2 e
l t
1
= c1 e
l t
1
v1 C c2 e
v2
v2 C t v1 .
Summary!!!! In the case of non-diagonalizable A2 # 2 ,
2
A K l I = l K l1
with l 1 real, and dim El
1
= 1. Find an eigenvector v1 and then a generalized eigenvector v2 , i.e. find
solutions to
A K l 1 I v1 = 0
A K l 1 I v2 = v1 .
Then the the two-dimensional solution space to x#= A x is given by linear combinations
l t
1
x t = c1 e
l t
1
v1 C c2 e
t v1 C v2 .
l t
1
Exercise 2) Verify directly that in the case above, the function e
x#= A x
t v1 C v2 does indeed solve
Exercise 3) Solve the linear system of DE's
x1 # t
x2 # t
=
1
x1
K4 0
x2
4
.