pure maths #92P1BQ13

From b(i),
given : A^2-tr(A)A=-(detA)I----*
b(ii), If tr(A^2)=0 and tr(A)=0
By (*) ,
A^2=-(detA)I
tr(A^2)=tr[-(detA)I]
tr(A^2)=(-detA)tr(I)
(-detA)(1+1)=0
-2detA=0
detA=0
A is singular
A^2=-(detA)I=0
(c)
Let A=(ST-TS)
So , we suffice to show tr(A^2)=tr(A)=0 and then A^2=0
1.checking tr(A)
tr(A)
=tr(ST-TS)
=tr(ST+(-1)(TS))
=tr(ST)-tr(TS) [by ai]
=tr(ST)-tr(ST) [by aii]
=0
2. checking tr(A^2)
tr(A^2)
=tr[(ST-TS)^2]
=tr[(ST-TS)(ST-TS)]
=tr[[(ST-TS)S]T-(ST-TS)TS]------**
=tr[[(ST-TS)S]T-TS(ST-TS)] [by aii]
=tr[[(ST-TS)S]T]-tr[(T)[S(ST-TS)]] [by ai]
=tr[S(ST-TS)T]-tr[(S(ST-TS)T] [by part c assumption]
=0
[For **, don't write as tr[[(ST-TS)(TS)-(ST-TS)TS]] in the next step as it's already reminded by aiii that it's wrong]
Hence , A^2=(ST-TS)^2=0

2009-03-23 19:22:32 補充：
For the last sentence , not"wrong" , should be "not necessarily true" ...