User:J D Bowen/Math725 HW9

1) Observe that $\langle T(a), \vec{w} \rangle = \langle a\vec{v}, \vec{w} \rangle = a\langle \vec{v}, \vec{w} \rangle= a\vec{v}\overline{\vec{w}}=a \overline{\vec{w}}\vec{v} = a \overline{\vec{w}\overline{\vec{v}} } = \langle a, \vec{w}\overline{\vec{v}}  \rangle =\langle a, \langle \vec{w},\vec{v} \rangle  \rangle=\langle a, S(\vec{w}) \rangle \ $.

So $T^H = S \ $.

2) a) Suppose $P\vec{v}=\vec{v} \ $. Then $P\vec{v}\in\text{im}(P) \implies \vec{v}\in\text{im}(P) \ $. Now suppose $\vec{v}\in\text{im}(P) \ $. Then $\exists\vec{u}:P\vec{u}=\vec{v} \ $. Since $P \ $ is a projection, $P^2\vec{u}=P\vec{u}=\vec{v}, \ $ but $P^2 \vec{u}= P(P\vec{u})=P\vec{v} \ $. Hence $P\vec{v}=\vec{v} \ $ and so $\vec{v}\in\text{im}(P) \ $.

b)

c) Suppose $P \ $ is an orthogonal projection, so that $P^2= P \ $ and $P\vec{v}-\vec{v} \perp \text{im}(P) \ $.

We have $\vec{v}\in\text{ker}(P) \iff \vec{0} - \vec{v} \perp \text{im}(P) \iff \vec{v}\perp\text{im}(P) \ $.

Since $\text{ker}(P)+\text{im}(P)=V \ $, we have $\text{ker}(P)=\text{im}(P)^\perp \ $.