4.3 Diagonal Form of a Matrix 574.3.2 Repeated Eigen Values with Full Kernels
In this case, (recall that d(s) = Yii=i(s ~ A;)ni)i we have dimAf([A — Xil]) —
Tii,Vz. Thus, there exists n* linearly independent vectors in Af ([A — Xil]), each
of which is an eigen vector associated with A;, Vi.Ai •H- Vn,Vl2,...,ViniA 2 f* V 2 l,V22,---,V2n 2Afc *-t Vkl,Vk2,...,Vknk
Ur=i {v»i}?ii 1S linearly independent (Exercise). Thus, we have obtained
n linearly independent vectors, which constitute a basis. Consequently, we getS~^1 AS =AiExample 4.3.5A =3 1 -1
13-1
00 2d(s) = det(sl - A) =s-3 -1 1
-1 s-3 1
0 0 s-2= 0= (* - 3)^2 (s - 2) - (a - 2) = (a - 2)[(* - 3)^2 - 1]
= (* - 2)(s - 4)(a - 2) = (a - 2)^2 (s - 4).=> Ai = 2, ni = 2 and A2 = 4, 712 = 1.
A-\iI =1 1 -1
1 1 -1
00 0dim(Af([A - Aj/])) = 2.vii = (l)-l,0)r,«i2 = (0,l,l)T.
-1 1 -1
A-\ 2 I= 1-1-1
0 0-2