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−12 −7 ⎟
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7.8 Jordan Form
http://www.amazon.com/exec/obidos/ASIN/0898714540 591 Solution: Computing the eigenvalues (which is the hardest part) reveals two
distinct eigenvalues λ1 = 2 and λ2 = −1, so there are two Jordan segments in
0
the Jordan form J = J(2) J(−1) . Computing ranks ri (2) = rank (A − 2I)i
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and ri (−1) = rank (A + I)i until rk ( ) = rk+1 ( ) yields r1 (2) = rank (A − 2I) = 4, r1 (−1) = rank (A + I) r2 (2) = rank (A − 2I) = 3, r2 (−1) = rank (A + I) r3 (2) = rank (A − 2I) = 2, r4 (2) = rank (A − 2I) = 2, It is illegal to print, duplicate, or distribute this material
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3
4 = 4,
2 = 4, D
E so k1 = index (λ1 ) = 3 and k2 = index (λ2 ) = 1. This tells us that the largest
Jordan block in J(2) is 3 × 3, while the largest Jordan block in J(−1) is 1 × 1
so that J(−1) is a diagonal matrix (the associated eigenvalue is semisimple
whenever this happens). Furthermore, T
H ν3 (2) = r2 (2) − 2r3 (2) + r4 (2) = 1 =⇒ one 3 × 3 block in J(2), ν2 (2) = r1 (2) − 2r2 (2) + r3 (2) = 0 =⇒ no 2 × 2 blocks in J(2), ν1 (2) = r0 (2) − 2r1 (2) + r2 (2) = 1 =⇒ one 1 × 1 block in J(2), IG
R ν1 (−1) = r0 (−1) − 2r1 (−1) + r2 (−1) = 2 =⇒ two 1 × 1 blocks in J(−1).
⎛ 2 1 0 ⎜0 2 1
Therefore, J(2) = ⎝ 0 0 2 ⎞ 0
0⎟
0⎠ Y
P
0 0 0 2 ⎛ O
C
J= J(2)
0
0
J(−1) ⎜
⎜
⎜
⎜
=⎜
⎜
⎜
⎜
⎝ and J(−1) = −1 0 0 −1 210
021
002 0
0
0 0
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0 0 0 0 2 0 0 0 0 0 −1 0 0 0 0 0 so that 0
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−1 The above example suggests that determining the Jordan form for An×n
is straightforward, and perhaps even easy. In theory, it is—just ﬁnd σ (A) , and
calculate some ranks. But, in practice, both of these tasks can be diﬃcult. To
begin with, the rank of a matrix is a discontinuous function of its entries, and r...

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