Exercise 3.F.28

Answers

Proof. Similarly to Exercise 27, we have

\begin{aligned} range T & = \\ v \in V : ψ (v) = 0 for every ψ \in {(range T)}^{0} & = & v & \in & V & : & ψ & ( & v & ) & = & 0 & for every & ψ & \in & null & T^{'} & = & v & \in & V & : & ψ & ( & v & ) & = & 0 & for every & ψ & \in & span & ( & φ & ) & = & v & \in & V & : & φ & ( & v & ) & = & 0 & = & v & \in & V & : & v & \in & null & φ & = & null & φ \end{aligned}

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2017-10-06 00:00