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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, APRIL 27 Repeated Eigenvalues
Jordan Canonical Form. We explain the general theory for when we can diagonalize a 2 × 2 matrix A.
Say that we have eigenvalues r1 and r2 .
i. If r1 = r2 i.e., we have distinct eigenvalues, then there exist 2 × 2 matrices
ξ
ξ
r
0
T = 11 12
and
D= 1
such that
T−1 A T = D.
ξ21 ξ22
0 r2
ii. If r1 = r2 i.e., we have repeated eigenvalues, then there exist 2 × 2 matrices
ξ11 η1
r1 c
T=
and
J=
such that
T−1 A T = J.
ξ21 η2
0 r1 Here, either c = 0 or c = 1. In fact, c = 0 if and only if A = r1 I is a diagonal matrix.
Note that A is diagonalizable if and only if either (1) A has distinct eigenvalues or (2) A = r I is a scalar
multiple of the identity matrix. The matrices D and J are called the Jordan canonical form for A.
We sketch the proof of this result. Denote the eigenvectors of r1 and r2 as
ξ
ξ
ξ (1) = 11
and
ξ (2) = 12 .
ξ21
ξ22
We have three cases to consider:
Case #1: r1 = r2 . We saw in the previous lecture that
ξ
ξ
rξ
r2 ξ12
T = 11 12
=⇒
A T = 1 11
= TD
ξ21 ξ22
r1 ξ21 r2 ξ22
Case #2: r1 = r2 and A = r1 I. Choose the matrix
10
r
T=
=⇒
T−1 A T = 1
01
0
Case #3: r1 = r2 and A = r1 I. Choose a nonzero vector
η
η= 1
via the relation
η2
Since A η = ξ (1) + r1 η , we set
ξ
η1
T = 11
=⇒
ξ21 η2 AT =
r1 ξ11
r1 ξ21 =⇒
0
=J
r1 T−1 A T = D. where c = 0. (r1 I − A) η = −ξ (1) .
ξ11 + r1 η1
= TJ
ξ21 + r1 η2 =⇒ T−1 A T = J; where c = 1. Since (r1 I − A) ξ (1) = 0, we say that ξ (1) is an eigenvector of A. Similarly, since (r1 I − A) η =
−ξ (1) we say that η is a generalized eigenvector of A. This proof shows that when A has repeated eigenvalues,
we can always choose T such that c = 0 or 1.
Example. Consider the following 2 × 2 matrix:
A=
1
1
−1
.
3 Recall that we have the eigenvalue r1 = r2 = 2. Deﬁne the vectors
1
0
ξ (1) =
and η =
=⇒
(r1 I − A) ξ (1) = 0 and
−1
−1
Hence we choose the 2 × 2 matrix
ξ11 η1
1
T=
=
ξ21 η2
−1
0
−1 =⇒
1 J=T −1 (r1 I − A) η = −ξ (1) .
2
AT =
0
1
.
2 ...
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 Spring '09
 EdrayGoins

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