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 $$\text{ker}(P)=\text{im}(P)^\perp \ $$.