I Proof of ##F## is an orthogonal projection if and only if symmetric

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Summary
This is a proof of a linear transformation ##F## on an inner product space ##V## being an orthogonal projection if and only if ##F## is a projection and symmetric.
The given definition of a linear transformation ##F## being symmetric on an inner product space ##V## is

##\langle F(\textbf{u}), \textbf{v} \rangle = \langle \textbf{u}, F(\textbf{v}) \rangle## where ##\textbf{u},\textbf{v}\in V##.​

In the attached image, second equation, how is the second equality justified? That is, ##\langle F(\textbf{v}), F(\textbf{v}) \rangle = \langle \textbf{v}, F(F(\textbf{v})) \rangle##. For projections in general, ##F=F^2##, but why does ##F(\textbf{v})=\textbf{v}## for ##\textbf{v} \in (\text{im} \ F)^{\perp}##


IMG_3099.jpg
 

Math_QED

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Summary: This is a proof of a linear transformation ##F## on an inner product space ##V## being an orthogonal projection if and only if ##F## is a projection and symmetric.

The given definition of a linear transformation ##F## being symmetric on an inner product space ##V## is

##\langle F(\textbf{u}), \textbf{v} \rangle = \langle \textbf{u}, F(\textbf{v}) \rangle## where ##\textbf{u},\textbf{v}\in V##.​

In the attached image, second equation, how is the second equality justified? That is, ##\langle F(\textbf{v}), F(\textbf{v}) \rangle = \langle \textbf{v}, F(F(\textbf{v})) \rangle##. For projections in general, ##F=F^2##, but why does ##F(\textbf{v})=\textbf{v}## for ##\textbf{v} \in (\text{im} \ F)^{\perp}##


View attachment 250815
Apply the definition of symmetric linear map you quoted.
 
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Hi, for the second equality you've got : ##||F(v)||^2 = <v, F(F(v))>## (because ##F## is symmetric) and this equate ##0## since ##v \in Im(F)^{\perp}## and ##F(F(v)) \in Im(F)##. Where is the problem?

Perhaps I didn't understand the question.
 

Delta2

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Apply the definition of F you quoted for ##u=v##, ##v=F(v)## (those are replacement equations, not direct equations, i.e ##v## isn't something special such that ##v=F(v)##.
 

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