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Linear Algebra - Sebastian Pancratz

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n = 4 (λ−t) 4 4 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1 <br />

<br />

⎛<br />

⎞ ⎛<br />

⎞ ⎛<br />

⎞<br />

λ 1<br />

λ 1<br />

λ<br />

⎟<br />

1⎠<br />

λ<br />

⎜<br />

⎝<br />

(λ − t) 4<br />

⎛<br />

λ 1<br />

⎜<br />

⎝<br />

λ<br />

(λ − t) 2<br />

λ<br />

λ<br />

⎞<br />

⎟<br />

⎠<br />

⎜<br />

⎝<br />

λ 1<br />

(λ − t) 3<br />

⎛<br />

λ<br />

⎜ λ<br />

⎝<br />

λ − t<br />

λ 1<br />

λ<br />

n((α − λι) r ) r <br />

<br />

⎛<br />

2 0 0<br />

⎞<br />

0<br />

⎜<br />

A = ⎜3<br />

⎝0<br />

2<br />

0<br />

0<br />

2<br />

−2 ⎟<br />

0 ⎠<br />

0 0 2 2<br />

χA(t) = (2 − t) 4 <br />

⎛<br />

0 0 0<br />

⎞<br />

0<br />

⎜<br />

A − 2I = ⎜3<br />

⎝0<br />

0<br />

0<br />

0<br />

0<br />

−2 ⎟<br />

0 ⎠<br />

0 0 2 0<br />

λ<br />

λ<br />

λ<br />

⎟<br />

⎠<br />

⎞<br />

⎟<br />

⎠<br />

⎜<br />

⎝<br />

λ 1<br />

λ<br />

(λ − t) 2<br />

⎟<br />

λ 1⎠<br />

λ<br />

(A − 2I) 2 ⎛<br />

0 0 0<br />

⎞<br />

0<br />

⎜<br />

= ⎜0<br />

⎝0<br />

0<br />

0<br />

−4<br />

0<br />

0 ⎟<br />

0⎠<br />

0 0 0 0<br />

A mA(t) = (2 − t) 3 n(A − 2I) = 2 <br />

A <br />

⎛<br />

2 1 0<br />

⎞<br />

0<br />

⎜<br />

JNF = ⎜0<br />

⎝0<br />

2<br />

0<br />

1<br />

2<br />

0 ⎟<br />

0⎠<br />

0 0 0 2<br />

.<br />

v3 ∈ ker(A − 2I) 2 v3 = (0, 0, 1, 0) T A − 2I<br />

⎛ ⎞<br />

0<br />

⎜<br />

v3 = ⎜0<br />

⎟<br />

⎝1⎠<br />

0<br />

↦→ v2<br />

⎛ ⎞<br />

0<br />

⎜<br />

= ⎜0<br />

⎟<br />

⎝0⎠<br />

2<br />

↦→ v1<br />

⎛ ⎞<br />

0<br />

⎜<br />

= ⎜−4<br />

⎟<br />

⎝ 0 ⎠ ↦→ 0.<br />

0<br />

v4 v4 = (2, 0, 0, 3) T

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