Matrix j in 784 is called the jordan form for a the

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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 ⎪ ⎪ ⎭ ⎩ 1 Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 584 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 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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This document was uploaded on 03/06/2014 for the course MA 5623 at City University of Hong Kong.

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