Operator T∈End(F2) i.e T:F2⟶F2 such that T(x,y)=(y,0)
Let,
U={(x,0):x∈F} Note that, for any u∈U ,thus u=(x1,0) for some x1∈F
T(u)=(0,0)(∵ By definition of T) which implies T(u)∈U ,Hence T is invariant under U .
Also note that u was arbitrary and u∈U⟹u∈F2 but T∣U:U⟶U ,thus
T∣U(u)=T(u)=0⟹T∣U=0
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