I'm trying to prove this:
Let $X,Y$ Hilbert spaces and $A\in L(X,Y)$ such that $R(A)=Y$. Also $V=\ker(A)$. Given the operator of orthogonal projection $P:X\rightarrow V$ we define $ B: Y \rightarrow V^\perp$, such that $B(y)=x-P(x)$ for all $y\in Y$ with $x\in X$ where $A(x)=y$. Prove that $B$ is surjective and bounded..
I have proved that the Operator $B$ is well defined, linear and injective, but i can't prove that $B$ is surjective
Thanks :)
That $B: Y \to V^\perp$ is surjective just means that for every $z\in V^\perp$, there exists $y\in Y$ such that $B(y)=x$. To show that such a value of $y$ exists, just let $y = A(x).$ Then $B(y) = x - P(x).$ If we can show $P(x)=0$, then that says $B(y)=x$ and we're done. Since $x\in V^\perp,$ we have $P(x)=0$ by the definitions of orthogonal projection and of $V\mapsto V^\perp.$