✔ 最佳答案
(a) By the information given, Tx = λx. So T-1(Tx) = T-1(λx)
T-1(λx) = x. Sub. x = y/λ
T-1(y) = y/λ
This prove that the eigenvalue of the inverse T-1 is 1/λ
(b) Since ST=TS. If the image space S(V) of S is not T-invariant, then there will have one element x such that it is in S(V) but the image TS(x) is not in S(V).
If this is the case, then TS(x) ∉ S(V).
But TS(x) = ST(x). by the expression of left hand side, we see that the image of TS(x) should be in S(V) (because T(x) ∈ V), which contradicts our previous result. So, S(V) should be T-invariant.