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Linear Algebra - Sebastian Pancratz

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m = mj Cl = J(nl, λ) <br />

aj<br />

aj = mj<br />

i=1 ni gj = mj ej = n1<br />

n1 ≥ n1 ≥ · · · nm > 0<br />

n1 + n2 + · · · + nm = 0<br />

C <br />

λ1, . . . , λk<br />

<br />

V Wj = N(α − λjι) aj V =<br />

k j=1 Wj B = k j=1 Bj Bj Wj <br />

⎛<br />

B1<br />

⎜<br />

[α]B = ⎝<br />

⎞<br />

0<br />

⎟<br />

⎠ .<br />

0 Bk<br />

pj(t) = (λj −t) −aj k r=1 (λr −t) ar qj k Wj = Im(hj(α))<br />

j=1 pjqj = 1<br />

α(Wj) ⊂ Wj Wj α| ∈ End(Wj)<br />

Wj<br />

λ = λj V = Wj n = aj (α − λι) n = 0 α − λι <br />

V <br />

⎛<br />

λ<br />

⎜<br />

⎜1<br />

⎜<br />

⎝<br />

⎞<br />

0<br />

⎟<br />

⎠<br />

0 1 λ<br />

.<br />

λ v1, . . . , vm <br />

α−λι v1 ↦→ v2 ↦→ . . . ↦→ vm ↦→ 0 <br />

<br />

n = 3 <br />

<br />

⎛ ⎞<br />

⎛ ⎞ ⎛ ⎞<br />

⎝<br />

λ1<br />

λ2<br />

λ3<br />

⎠<br />

⎝<br />

λ1<br />

λ2 1 ⎠<br />

(λ1 − t)(λ2 − t)(λ3 − t) (λ1 − t)(λ2 − t) 2<br />

(λ1 − t)(λ2 − t)(λ3 − t) (λ1 − t)(λ2 − t) 2<br />

⎛ ⎞<br />

λ<br />

⎛ ⎞<br />

λ 1<br />

⎝ λ ⎠<br />

⎝ λ ⎠<br />

λ<br />

λ<br />

(λ − t) 3<br />

(λ − t) 3<br />

λ − t (λ − t) 2<br />

λ2<br />

⎝<br />

λ1<br />

λ2<br />

λ2<br />

⎠<br />

(λ1 − t)(λ2 − t) 2<br />

(λ1 − t)(λ2 − t)<br />

⎛ ⎞<br />

λ 1<br />

⎝ λ 1⎠<br />

λ<br />

(λ − t) 3<br />

(λ − t) 3

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