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

[schniefen]

TL;DR

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 $\textbf{v} \in (\text{im} \ F)^{\perp}$

 

[member 587159]

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.

 

[Geometry]

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]

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 $v=F(v)$.