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I hope someone can check the solution for me.

Here is the problem:

Let $V=V_1\oplus V_2$, $f$ is the projection of $V$ onto $V_1$ along $V_2$( i.e. if $v=v_1+v_2, v_i\in V_i$ then $f(v)=v_1$). Prove that $f$ is self-adjoint iff $<V_1,V_2>=0$

my solution is this:

proof:"$\Rightarrow$"let $v_1\in V_1, v_2\in V_2$,

then $<f(v_1),v_2>=<v_1,f*(v_2)>=<v_1,f(v_2)>$, since if $f$ is self-adjoint,

then $f(v_2)=0, f(v_1)=v_1$, it follows that $<v_1,v_2>=<v_1,0>=0$,

hence $<v_1,v_2>=0$

"$\Leftarrow$" let $v_1\in V_1, v_2\in V_2$,

$<f(v_1),v_2>=<v_1,v_2>=0$, since $<v_1,v_2>=0$

$<v_1,f(v_2)>=<v_1,0>=0$

hence $<f(v_1),v_2>=<v_1,f(v_2)>$, $f$ is self adjoint.

It seems like something is wrong with my proof, but I really don't know. Hope someone can check it.

Thanks