Math 2280-001
Mon Mar 30
5.5 Linear systems x#= A x for which A is not diagonalizable.
We will spend a large part of the lecture finishing Friday's notes on section 5.5.
Using the Jordan canonical form of a non-diagonalizable matrix, as is carried out for a 2 # 2 matrix in
Friday's notes, one is led (after omitted steps) to the following solution algorithm. Let J be the Jordan
form of A, so that for a invertible matrix S whose columns are generalized eigenvectors,
A S = S J.
, For each 2 # 2 Jordan block in J
l 1
0 l
There are successive columns v1 , v2 in S making a chain of length 2:
A v1 = l v1
A v2 = l v2 C v1
i.e.
A K l I v1 = 0
A K l I v2 = v1 .
There are two corresponding independent solutions to the first order system x#= A x, namely
x1 t = el t v1
x2 t = el t v2 C t v1 .
, For each 3 # 3 Jordan block in J
l 1 0
0 l 1
0 0 l
There are three successive columns v1 , v2 , v3 in S, making a chain of length 3
A K l I v1 = 0
A K l I v2 = v1
A K l I v3 = v2
There are three corresponding independent solutions to the first order system x#= A x, namely
x1 t = el t v1
x2 t = el t v2 C t v1 .
x3 t = el t v3 C t v2 C
, The procedure is analogous for larger Jordan blocks.
t2
v .
2 1
Exercise 1: Use technology and the methods above to find the general solution to
x1 # t
x2 # t
K2
0
1
0
K3
0
K1
0
K4
=
x3 # t
.
The following output will let you answer this question:
> with LinearAlgebra :
K2 0 1
> A d 0 K3 0 ;
K1 0 K4
K2 0 1
0 K3 0
A :=
K1 0 K4
(1)
> B d A K l$IdentityMatrix 3 ;
factor Determinant B ; # factored characteristic polynomial
B :=
K2 K l
0
1
0
K3 K l
0
K1
0
K4 K l
K3 C l
>
Thus p l =K l C 3
3
3
(2)
and l =K3 has algebraic multiplicity 3.
> Eigenvectors A ;
K3
K3 ,
K3
0 K1 0
1 0 0
0 1 0
(3)
The eigenspace is only 2-dimensional, however, so the defect is 1. The Jordan form of A must have a
2 # 2 block and a 1 # 1 block. We can check this:
> JordanForm A ;
K3 1 0
0 K3 0
0 0 K3
(4)
> C d A C 3$IdentityMatrix 3 ;
C 2; # should be zero. why?
1 0 1
0 0 0
C :=
K1 0 K1
0 0 0
0 0 0
0 0 0
(5)
Solution: For the 2 # 2 block I may take v1 the first column of C and v2 = 1, 0, 0
T
since then
A C 3 I v2 = v1 . For the second eigenvector v3 we use 0, 1, 0 T, from the "Eigenvectors" command
output. If I put v1 , v2 , v3 into the columns of a matrix S, then I should recover the Jordan form as SK1 A S:
1 1 0
> Sd 0 0 1 ;
K1 0 0
SK1.A.S; #should be J..note we multiply matrices using a period.
1 1 0
0 0 1
S :=
K1 0 0
K3 1 0
0 K3 0
0 0 K3
(6)
General solution to x#= A x is
x1 t
x2 t
x3 t
1
= c1
eK3 t
0
K1
1
C c2
eK3 t
0
0
1
Ct
0
K1
0
eK3 t
C c3
1
.
0
Warning: Since the eigenspace is 2Kdimensional, and because there were a lot of choices for the chain of
length 2 for the 2 # 2 Jordan block, not all correct general solutions will immediately appear to be
equivalent....
> with DEtools :
> dsolve x# t =K2$x t C z t , y# t =K3$y t , z # t =Kx t K 4$z t , x t , y t , z t ;
x t = eK3 t _C1 C _C2 t , y t = _C3 eK3 t , z t = KeK3 t _C1 C _C2 t K _C2
(7)
>
So Maple's version actually agrees with ours in this case, without any changes.
, There is a beautiful example of using this theory, in order to understand a coupled damped 2-car "train",
Example 6 in section 5.5 If we have time we may discuss this on Wednesday, although we may also want
to review for the midterm on Friday.
Universal product rule for differentiation: Recall the 1-variable product rule for differentiation for a
function of a single variable t, based on the limit definition of derivative. We'll just repeat that discussion,
but this time for any product "∗" that distributes over addition, for scalar, vector, or matrix functions. We
also assume that for the product under consideration, scalar multiples s behave according to
s f)g = sf )g = f) sg .
We don't assume that f)g = g)f so must be careful in that regard. Here's how the proof goes:
1
Dt f)g t d lim
f t C Dt )g t C Dt K f t )g t .
Dt / 0 Dt
We add and subtract a middle term, to help subsequent algebra:
1
= lim
f t C Dt )g t C Dt K f t C Dt )g t C f t C Dt )g t Kf t )g t
Dt / 0 Dt
1
1
= lim
f t C Dt )g t C Dt K f t C Dt )g t C
f t C Dt )g t Kf t )g t
Dt / 0 Dt
Dt
We assume that multiplication by ∗ distributes over addition:
1
1
= lim
f t C Dt ) g t C Dt Kg t C
f t C Dt K f t )g t .
Dt / 0
Dt
Dt
The sum rule for limits and rearranging the scalar factor let's us rearrange as follows:
1
1
= lim f t C Dt )
g t C Dt Kg t C lim
f t C Dt K f t )g t .
Dt / 0
Dt / 0 Dt
Dt
Differentiable functions are continuous, so we take limits and get:
Dt f)g t = f t )g# t C f# t )g t .
.
The proof above applies to
scalar function times scalar function (Calc I)
scalar function time vector function (Calc III)
dot product or cross product of two functions (Calc III)
scalar function times matrix function (our class)
matrix function times matrix or vector function (our class)
This proof does not apply to composition f + g t d f g t , because composition does not generally
distribute over addition, and this is why we have the chain rule for taking derivatives of composite
functions.
A