Given a separable Hilbert space $H$ and its subspace $L$. Let $B_1$,....,$B_n$ be bounded operators on $H$ and $P_L$ be the projection onto $L$. If there exist positive real numbers $\lambda_1$, ...,$\lambda_n$ such that $P_L{B_r}^*B_sP_L=\delta_{rs}\lambda_rP_L$. Then I have to show that there are partial isometries $U_1$, ... , $U_n$ that map $L$ onto mutually orthogonal subspaces $L_s$ of $H$ and
\begin{equation*} \lambda_s^{-1/2}B_sP_L=U_sP_L \end{equation*}
holds. What is a partial isometry? And how to show this equality? Could anyone please help me?
Note that if You define $$U_{s|L}:L\rightarrow L_s,\phi\mapsto \lambda_s^{-\frac{1}{2}}B_s\phi$$ You could by $$(\lambda_s^{-\frac{1}{2}}B_s\phi,\lambda_s^{-\frac{1}{2}}B_s\phi)=\lambda_s^{-1}(B_s\phi,B_s\phi)$$ $$=\lambda_s^{-1}(B_s^*B_s\phi,\phi)\underbrace{=}_{\phi\in L}$$$$\lambda_s^{-1}(P_LB_s^*B_s\phi,\phi)=\lambda_s^{-1}\lambda_s(\phi,\phi)=(\phi,\phi)$$ define a partial isometry if you say $U_{s|L^{\perp}}\equiv 0$.(I assume $L$ closed.) The equations show $U_s$ is an isometry on $L$ which is the orthogonal complement of its kernel.(Note that $L\cap ker U_s=0$ because all $\lambda_s$ are positive.)
$U_s$ fulfills the equality You wrote and for $L_r\neq L_s$ a similar argument shows $$(\lambda_s^{-\frac{1}{2}}B_s\phi,\lambda_r^{-\frac{1}{2}}B_r\phi)=...=\lambda_s^{-\frac{1}{2}}\lambda_r^{-\frac{1}{2}}\delta_{sr}\lambda_r(\phi,\phi)=0$$ and thus $L_r\perp L_s$.