行列の対角化解答
1. (1) A= (2-2)
7
16
3
BgA(t) = (t-1) (t~4)
固有値入=1,4
BAHW (1:A) = span [ (1)]
W (4:A) = span [ (²)
dim W (1A) + dim W (4:A)`
/+/
2
Aは対角化可能
P = ( ₁² ) etice P²= (72)
PAP = (14)
(2) A = (
-30
A = ( 13² - 12)
5-12
B\\ J₁ (t) = (t+2)(t−3)
固有値入=12,3
DAGH W(-2A) = span [(?)]
W(3;A) = span [ (³)]
din W (2=A) + dim W(3=A)
/+/
2
3. A FÁE
2 3
P= ( ² ; ) Lack P²= (^_^)
とおくと
(713)
PAP
(3)
=
A =
20
(23)
03
(4)
2-12
BþÑÃI) JA (†) = (t+1) (t-1)²
固有値入=-1,1
222-1.
dim W(-1A) + dim W(1-A)
= / +/
= 2 < 3
W(-1: A) = span [ (+;)]
W(1=A) = span [(!)]
・Aは対角化可能でない
A =
6
342
-8-4-3
固有多項式gA(t)=(りる
固有値入=1, 3
2
