let [MATH](E,\langle\cdot,\cdot\rangle)[/MATH] an euclidean vector space with [MATH]\dim(E)=2n[/MATH]. Let F and G two non-orthogonal linear subspaces of E with [MATH]\dim(F)=\dim(G)=n[/MATH]. Let [MATH]p_{_{F}}[/MATH] be the orthogonal projection on F and [MATH]p_{_{G}}[/MATH] be the orthogonal projection on G. Prove that [MATH]p_{_{F}}\circ p_{_{G}}[/MATH] has at least one strictly positive eigenvalue.
The characteristic polynomial of [MATH]p_{_{F}}\circ p_{_{G}}[/MATH] and [MATH]p_{_{G}} \circ p_{_{F}}\circ p_{_{G}}[/MATH] are the same since [MATH]p_{_{G}}^2 = p_{_{G}}[/MATH].
Since [MATH]p_{_{G}}[/MATH] and [MATH]p_{_{F}}[/MATH] are orthogonal projections, they are self-ajoint operators.
Let [MATH](x,y)\in E^2[/MATH][MATH] \langle (p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,y\rangle = \langle ( p_{_{F}}\circ p_{_{G}})(x) ,p_{_{G}}(y)\rangle = \langle p_{_{G}}(x) ,(p_{_{F}} \circ p_{_{G}})(y)\rangle = \langle x ,(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})(y)\rangle[/MATH]So [MATH]p_{_{G}} \circ p_{_{F}} \circ p_{_{G}}[/MATH] is a self-ajoint operator
Ad absurdum, let's suppose that [MATH]p_{_{G}} \circ p_{_{F}} \circ p_{_{G}}[/MATH] has only negative eigenvalues, ie [MATH]-(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})[/MATH] is a positive self-adjoint operator ie [MATH]\forall x \in E, \langle -(p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,x\rangle \geq 0[/MATH]
if [MATH] F \cap G \neq \lbrace{0}\rbrace[/MATH] , let [MATH]x\in (F \cap G)\setminus \lbrace{0}\rbrace[/MATH][MATH]\langle -(p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,x\rangle = -\langle x,x \rangle < 0[/MATH]hence the contradiction because [MATH]-(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})[/MATH] is a positive self-adjoint operator
I could not find the case where [MATH]F \cap G = \lbrace{0}\rbrace[/MATH]
The characteristic polynomial of [MATH]p_{_{F}}\circ p_{_{G}}[/MATH] and [MATH]p_{_{G}} \circ p_{_{F}}\circ p_{_{G}}[/MATH] are the same since [MATH]p_{_{G}}^2 = p_{_{G}}[/MATH].
Since [MATH]p_{_{G}}[/MATH] and [MATH]p_{_{F}}[/MATH] are orthogonal projections, they are self-ajoint operators.
Let [MATH](x,y)\in E^2[/MATH][MATH] \langle (p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,y\rangle = \langle ( p_{_{F}}\circ p_{_{G}})(x) ,p_{_{G}}(y)\rangle = \langle p_{_{G}}(x) ,(p_{_{F}} \circ p_{_{G}})(y)\rangle = \langle x ,(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})(y)\rangle[/MATH]So [MATH]p_{_{G}} \circ p_{_{F}} \circ p_{_{G}}[/MATH] is a self-ajoint operator
Ad absurdum, let's suppose that [MATH]p_{_{G}} \circ p_{_{F}} \circ p_{_{G}}[/MATH] has only negative eigenvalues, ie [MATH]-(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})[/MATH] is a positive self-adjoint operator ie [MATH]\forall x \in E, \langle -(p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,x\rangle \geq 0[/MATH]
if [MATH] F \cap G \neq \lbrace{0}\rbrace[/MATH] , let [MATH]x\in (F \cap G)\setminus \lbrace{0}\rbrace[/MATH][MATH]\langle -(p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,x\rangle = -\langle x,x \rangle < 0[/MATH]hence the contradiction because [MATH]-(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})[/MATH] is a positive self-adjoint operator
I could not find the case where [MATH]F \cap G = \lbrace{0}\rbrace[/MATH]