Lecture Notes in Differential Equations - Bruce E. Shapiro

318 LESSON 31. LINEAR SYSTEMS
Proof. We have show previously that each function in equations (31.113) is
a solution of y ′ = Ay. Since the eigenvectors are linearly independent, the
solutions are also linearly independent. It follows that the solutions form a
fundamental set.
Generalized Eigenvectors
As a corollary to theorem 31.10, we observe that if A is diagonalizable
(i.e. it has n linearly independent eigenvectors) then the general solution
of y ′ = Ay is
n∑
y = C i v i e λit (31.114)
i=1
If the matrix is not diagonalizable, then to find the fundamental matrix we
must first find the Jordan form J of A, because
e A = Ue J U −1 (31.115)
where J = U −1 AU for some matrix U that we must determine. If A were
diagonalizable, the columns of U would be the eigenvectors of A. Since the
system is not diagonalizable there are at least two eigenvectors that share
the same eigenvalue.
Let J = U −1 AU be the Jordan Canonical Form of A. Then since A = UJU −1 ,
we can rearrange the system of differential equations y ′ = Ay as
Let z = U −1 y. Then
y ′ = Ay = UJU −1 y (31.116)
Uz ′ = y ′ = UJz (31.117)
Multiplying through by U −1 on the left, we arrive at the Jordan form of
the differential equation,
z ′ = Jz (31.118)
where J is the Jordan Canonical Form of A. Hence J is block diagonal,
with each block corresponding to a single eigenvalue λ i of multiplicity m i .
Writing
⎛
⎞
B 1 0 · · · 0
⎛
z ′ = Jz =
0 B 2 .
⎜
⎝
. ⎟ ⎜
. .. 0 ⎠ ⎝
0 · · · 0 B k
z 1
z 2
.
z k
⎞
⎟
⎠
(31.119)