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**Unformatted text preview: **r3 = 0, so the index of L is k = 3, and
the number of 3 × 3 blocks = r2 − 2r3 + r4 = 1,
the number of 2 × 2 blocks = r1 − 2r2 + r3 = 1,
the number of 1 × 1 blocks = r0 − 2r1 + r2 = 1. It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] Consequently, the Jordan form of L is
⎛ 0
⎜0
⎜
⎜0
⎜
N=⎜
⎜0
⎜
⎝0
0 10
01
00 0
0
0 0
0
0 00
00 0
0 1
0 ⎞
0
0⎟
⎟
0⎟
⎟
⎟.
0⎟
⎟
0⎠ 0 0 0 0 0 D
E T
H Notice that three Jordan blocks were found, and this agrees with the fact that
dim N (L) = 6 − rank (L) = 3. Determine P by following the procedure described in Example 7.7.3.
1. Since rank L2 = 1, any nonzero column from L2 will be a basis for
M2 = R L2 , so set y1 = [L2 ]∗1 = (6, −6, 0, 0, −6, −6)T .
2. To extend y1 to a basis for M1 = R (L) ∩ N (L), use Y
P IG
R ⎛ 1
⎜3
⎜
⎜ −2
B = [L∗1 | L∗2 | L∗3 ] = ⎜
⎜2
⎝
−5
−3 O
C 1
1
−1
1
−3
−2 ⎞
⎛
−2
6
5⎟
⎜ −6
⎟
⎜
0⎟
⎜0
⎟ =⇒ LB = ⎜
0⎟
⎜0
⎠
⎝
−1
−6
−1
−6 and determine a basis for N (LB) to be v1 = −1
2
0 ⎞
3
3
−3 −3 ⎟
⎟
0
0⎟
⎟,
0
0⎟
⎠
−3 −3
−3 −3 , v2 = −1
0
2 . Reducing [ y1 | Bv1 | Bv2 ] to echelon form shows that its basic columns
are in the ﬁrst and third positions, so {y1 , Bv2 } is a basis for M1 with
⎫
⎧⎛
⎞
6
⎪
⎪
⎪
⎪
⎪
⎪⎜ −6 ⎟
⎪
⎪
⎪
⎪⎜
⎟
⎬
⎨
⎜ 0⎟
S2 = ⎜
⎟ = b1
⎪
⎪⎜ 0 ⎟
⎪
⎪⎝
⎪
⎪
⎪
⎪ −6 ⎠
⎪
⎪
⎭
⎩
−6 Copyright c 2000 SIAM and ⎫
⎧⎛
⎞
⎪
⎪ −5
⎪
⎪
⎪
⎪⎜ 7 ⎟
⎪
⎪
⎪
⎪⎜
⎟
⎬
⎨
⎜ 2⎟
S1 = ⎜
⎟ = b2 .
⎪
⎪⎜ −2 ⎟
⎪
⎪⎝
⎪
⎪
⎠
⎪
⎪
3
⎪
⎪
⎭
⎩
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Chapter 7
Eigenvalues and Eigenvectors
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3. Now extend S2 ∪ S1 = {b1 , b2 } to a basis for M0 = N (L). This time,
B = I, and a basis for N (LB) = N (L) can be computed to be
⎞
2
⎜ −4 ⎟
⎟
⎜
⎜ −1 ⎟
v1 = ⎜
⎟,
⎜ 3...

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