[proofplan] The proof is a direct whitening computation. The choices $a_k=\Sigma_{11}^{-1/2}p_k$ and $b_k=\Sigma_{22}^{-1/2}r_k$ transform within-block covariances into Euclidean inner products of singular vectors. The cross-covariance becomes $p_k^\top C r_l$, which is diagonal because $p_k$ and $r_l$ are paired left and right singular vectors of $C$. [/proofplan] [step:Center the random vectors and record the covariance identities] Define the centered random vectors \begin{align*} \widetilde X &: \Omega \to \mathbb R^{d_1},& \widetilde X(\omega)&:=X(\omega)-\mathbb E[X],\\ \widetilde Y &: \Omega \to \mathbb R^{d_2},& \widetilde Y(\omega)&:=Y(\omega)-\mathbb E[Y]. \end{align*} Since $X$ and $Y$ are square-integrable, the real-valued random variables $U_k$ and $V_k$ are square-integrable. By the definitions of covariance matrices, \begin{align*} \operatorname{Cov}(U_k,U_l) &=a_k^\top \Sigma_{11}a_l,\\ \operatorname{Cov}(V_k,V_l) &=b_k^\top \Sigma_{22}b_l,\\ \operatorname{Cov}(U_k,V_l) &=a_k^\top \Sigma_{12}b_l. \end{align*} [/step] [step:Compute the covariance matrix of the $U_k$ variates] Using $a_k=\Sigma_{11}^{-1/2}p_k$ and $a_l=\Sigma_{11}^{-1/2}p_l$, and using that $\Sigma_{11}^{-1/2}$ is symmetric because $\Sigma_{11}$ is positive definite, we obtain \begin{align*} \operatorname{Cov}(U_k,U_l) &=a_k^\top\Sigma_{11}a_l\\ &=p_k^\top\Sigma_{11}^{-1/2}\Sigma_{11}\Sigma_{11}^{-1/2}p_l\\ &=p_k^\top p_l. \end{align*} The columns $p_1,\dots,p_m$ of $P$ are orthonormal, so $p_k^\top p_l=\delta_{kl}$. Hence \begin{align*} \operatorname{Cov}(U_k,U_l)=\delta_{kl}. \end{align*} [/step] [step:Compute the covariance matrix of the $V_k$ variates] Using $b_k=\Sigma_{22}^{-1/2}r_k$ and $b_l=\Sigma_{22}^{-1/2}r_l$, and using that $\Sigma_{22}^{-1/2}$ is symmetric because $\Sigma_{22}$ is positive definite, we obtain \begin{align*} \operatorname{Cov}(V_k,V_l) &=b_k^\top\Sigma_{22}b_l\\ &=r_k^\top\Sigma_{22}^{-1/2}\Sigma_{22}\Sigma_{22}^{-1/2}r_l\\ &=r_k^\top r_l. \end{align*} The columns $r_1,\dots,r_m$ of $R$ are orthonormal, so $r_k^\top r_l=\delta_{kl}$. Hence \begin{align*} \operatorname{Cov}(V_k,V_l)=\delta_{kl}. \end{align*} [/step] [step:Diagonalize the cross-covariance through the singular value decomposition] Using the definitions of $a_k$, $b_l$, and $C$, we compute \begin{align*} \operatorname{Cov}(U_k,V_l) &=a_k^\top\Sigma_{12}b_l\\ &=p_k^\top\Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2}r_l\\ &=p_k^\top C r_l. \end{align*} Since $C=PDR^\top$, and since $p_k$ and $r_l$ are the $k$th and $l$th columns of $P$ and $R$, respectively, \begin{align*} p_k^\top C r_l &=p_k^\top PDR^\top r_l\\ &=e_k^\top D e_l\\ &=\rho_k\delta_{kl}, \end{align*} where $e_1,\dots,e_m$ denotes the standard basis of $\mathbb R^m$. Therefore \begin{align*} \operatorname{Cov}(U_k,V_l)=\rho_k\delta_{kl}. \end{align*} [guided] The cross-covariance is the only place where the [singular value decomposition](/theorems/3071) enters. Starting from the covariance identity for scalar linear combinations, \begin{align*} \operatorname{Cov}(U_k,V_l) &=a_k^\top\Sigma_{12}b_l. \end{align*} Now substitute the canonical directions: \begin{align*} a_k&=\Sigma_{11}^{-1/2}p_k,& b_l&=\Sigma_{22}^{-1/2}r_l. \end{align*} Because $\Sigma_{11}^{-1/2}$ and $\Sigma_{22}^{-1/2}$ are symmetric positive definite square roots of the inverse covariance matrices, this gives \begin{align*} \operatorname{Cov}(U_k,V_l) &=p_k^\top\Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2}r_l\\ &=p_k^\top C r_l. \end{align*} Thus whitening has converted the original cross-covariance problem into a statement about the matrix $C$. Now use the singular value decomposition $C=PDR^\top$. The orthonormality of the columns means \begin{align*} p_k^\top P=e_k^\top,\qquad R^\top r_l=e_l, \end{align*} where $e_1,\dots,e_m$ is the standard basis of $\mathbb R^m$. Therefore \begin{align*} p_k^\top C r_l &=p_k^\top PDR^\top r_l\\ &=e_k^\top D e_l. \end{align*} Since $D=\operatorname{diag}(\rho_1,\dots,\rho_m)$, its $(k,l)$ entry is $\rho_k$ when $k=l$ and $0$ when $k\ne l$. Equivalently, \begin{align*} e_k^\top D e_l=\rho_k\delta_{kl}. \end{align*} Hence \begin{align*} \operatorname{Cov}(U_k,V_l)=\rho_k\delta_{kl}. \end{align*} [/guided] [/step] [step:Read off the variance and orthogonality conclusions] Setting $k=l$ in the within-block identities gives \begin{align*} \operatorname{Var}(U_k)&=\operatorname{Cov}(U_k,U_k)=1,\\ \operatorname{Var}(V_k)&=\operatorname{Cov}(V_k,V_k)=1. \end{align*} If $k\ne l$, then $\delta_{kl}=0$, so \begin{align*} \operatorname{Cov}(U_k,U_l)&=0,& \operatorname{Cov}(V_k,V_l)&=0,& \operatorname{Cov}(U_k,V_l)&=0. \end{align*} Finally, setting $k=l$ in the cross-covariance identity gives \begin{align*} \operatorname{Cov}(U_k,V_k)=\rho_k. \end{align*} These are exactly the asserted normalization, within-block orthogonality, and cross-block orthogonality relations. [/step]