Page 233 - DMTH502_LINEAR_ALGEBRA

Unit 21: The Jordan Form

Now Notes

0 0 0
a
I
(A 2 )(A I ) 3 0 0
ac 0 0
and thus A is similar to a diagonal matrix if and only if a = 0.

Example 3: Let

2 0 0 0
1 2 0 0
A
0 0 2 0
0 0 a 2

4
The characteristic polynomial for A is (x – 2) . Since A is the direct sum of two 2 × 2 matrices, it
2
is clear that the minimal polynomial for A is (x – 2) . Now if a = 0 or if a = 1, then the matrix A is
in Jordan form. Notice that the two matrices we obtain for a = 0 and a = 1 have the same
characteristic polynomial and the same minimal polynomial, but are not similar. They are not
similar because for the first matrix the solution space of (A – 2I) has dimension 3, while for the
second matrix it has dimension 2.

Example 4: Linear differential equations with constant coefficients provide a nice
illustration of the Jordan form. Let a ,..., a be complex numbers and let V be the space of all n
0 n–1
times differentiable functions f on an interval of the real line which satisfy the differential
equation
n
d f d n 1 f df
 a   a  a f  0
dx n n 1 dx n 1 1 dx 0
Let D be the differentiation operator. Then V is invariant under D, because V is the null space of
p(D), where
...
n
p = x + + a x + a
1 0
What is the Jordan form for the differentiation operator on V?
Let c ,..., c be the distinct complex roots of p:
1 k
r ... r
1 f
p = (x – c ) 1 (x – c ) k
k
r
Let V be the null space of (D – c I) i , that is, the set of solutions to the differential equation
i i
r
(D – c I) i f = 0
i
Then the primary decomposition theorem tells us that
V = V  ...  V
1 k
Let N be the restriction of D – c I to V . The Jordan form for the operator D (on V) is then
i i i
determined by the rational forms for the nilpotent operators N ,..., N on the spaces V ,..., V .
1 k 1 k
So, what we must know (for various values of c) is the rational form for the operator N = (D – cI)
on the space V , which consists of the solutions of the equation
c
r
(D – cI) f = 0

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