Proof. Simply note that the largest singular value is \[\max_{y \in \mathbb B}\max_{x \in \mathbb B}\langle P_F y, P_G x\rangle = \max_{y \in \mathbb B \cap F}\max_{x \in \mathbb B \cap G} \langle y,x\rangle = \theta_1\] where \(\theta_1\) is the largest principal angle. The iteration to the subsequent singular values is the same as the one used to find the principal angles. ◻

Proof. Put \[M = \begin{pmatrix}A & B \\ C & D\end{pmatrix}\] The maximum singular value of \(M\) is given by \(\Vert M \Vert_2\), which follows from the singular value decomposition. Then \[\begin{aligned} \Vert M \Vert_2^2 \geq \Vert M \vec v\Vert_2^2 = \Vert A \vec v \Vert_2^2 + \Vert C \vec v \Vert_2^2 \geq \Vert A \vec v \Vert_2^2 \end{aligned}\] Taking the supremum on both sides and using the definition \(\Vert A \Vert_2 = \sup_{\vec v \in \mathbb B}\Vert A\vec v \Vert\) proves the claim. ◻

Proof. Let \(M = \begin{pmatrix}X \\ Y\end{pmatrix}\). Choose unitaries \(U_1, V\) such that \(U_1^\dagger X V = C\), where \(C = \operatorname{diag}(c_1, \dots, c_k)\). Since \(X\) is a submatrix of \(M\) whose singular values are all 1, we have \(c_i \leq 1\). Then \[\begin{pmatrix}U_1^\dagger & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}X \\ Y\end{pmatrix}V = \begin{pmatrix}U_1^\dagger X V \\ YV\end{pmatrix} = \begin{pmatrix}C \\ YV\end{pmatrix}\] This transformation preserves the orthonormality of the columns of \(M\). We have \(M^\dagger M = C^2 + (YV)^\dagger (YV) = 1\). This implies that the columns of \(YV\) are orthogonal, and their norm is \(\Vert y_i \Vert^2 = 1-c_i^2\). Thus we can find a unitary matrix \(Q\) so that \(Q^\dagger YV = S\), where \(S = \operatorname{diag}(1-c_1^2, \dots, 1-c_k^2)\). Therefore \[\begin{pmatrix}U_1^\dagger & 0 \\ 0 & Q^\dagger\end{pmatrix}\begin{pmatrix}X \\ Y\end{pmatrix}V = \begin{pmatrix}U_1^\dagger X V \\ QYV\end{pmatrix} = \begin{pmatrix}C \\ S\end{pmatrix}\] This completes the proof. ◻

Proof. First, choose \(U_1, V_1\) so that \(U_1^\dagger X V_1 = C\). Since \(X\) is a submatrix of \(M\) whose singular values are all 1, we have \(C = \operatorname{diag}(c_1, \dots, c_k)\) where \(0 \leq c_i \leq 1\). Then we have \[\begin{pmatrix}U_1^\dagger & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}X & Y \\ Z & W\end{pmatrix} \begin{pmatrix}V_1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}U_1^\dagger X V_1 & U_1^\dagger Y \\ ZV_1 & W\end{pmatrix} = \begin{pmatrix}C & Y' \\ Z' & W\end{pmatrix}\] This transformation preserves the orthonormality of the rows and columns. Therefore \(\langle \vec z'_i, \vec z'_j \rangle = \delta_{ij}(1-c_i^2)\). This same equation is satisfied by the rows of \(Y'\). Therefore we can find unitaries \(R\) and \(L\) such that \(L^\dagger Z' = T\) and \(Y'R = -S\), where \(S = \operatorname{diag}(1-c_1^2, \dots, 1-c_k^2)\). As such,

\[\begin{pmatrix}U_1^\dagger & 0 \\ 0 & L^\dagger\end{pmatrix}\begin{pmatrix}X & Y \\ Z & W\end{pmatrix} \begin{pmatrix}V_1 & 0 \\ 0 & R\end{pmatrix} = \begin{pmatrix}U_1^\dagger X V_1 & U^\dagger_1YR \\ L^\dagger ZV_1 & L^\dagger WR\end{pmatrix} = \begin{pmatrix}C & S \\ -S & W'\end{pmatrix}\] Writing \(C = 1 \oplus C'\), where \(C'\) consists of \(c_i\) strictly less than 1, then we have \[\begin{pmatrix}C & S \\ -S & W'\end{pmatrix} = \begin{pmatrix}\mqty{1 & 0 \\ 0 & C'} & \mqty{0 & 0 \\ 0 & S'} \\ \mqty{0 & 0 \\ 0 & -S'} & \mqty{X_1 & X_2 \\ X_3 & X_4}\end{pmatrix}\] Taking the inner product of the last row with the second row gives \(S'X_4^\dagger-S'C' = 0\), where the unitarity of \(M\) guarantees that the inner product of rows is zero. Since \(S'\) is invertible (because \(1-c_i^2 > 0\)), we have \(X_4 = C'\), and by the orthonormality of the rows, \(X_2 = X_3 = 0\). This then implies that \(X_1 = 1\) by the same logic. Therefore we have \[\begin{pmatrix}\mqty{1 & 0 \\ 0 & C'} & \mqty{0 & 0 \\ 0 & S'} \\ \mqty{0 & 0 \\ 0 & -S'} & \mqty{X_1 & X_2 \\ X_3 & X_4}\end{pmatrix} = \begin{pmatrix}\mqty{1 & 0 \\ 0 & C'} & \mqty{0 & 0 \\ 0 & S'} \\ \mqty{0 & 0 \\ 0 & -S'} & \mqty{1 & 0 \\ 0 & C'}\end{pmatrix} =\begin{pmatrix}C & -S \\ S & C\end{pmatrix}\] Since \(0 \leq c_i \leq 1\), we can put \(c_i \equiv \cos(\theta_i)\) for some \(0 \leq \theta \leq \pi/2\), and \(s_i = 1-c_i^2 = \sin(\theta_i)\). This completes the proof. ◻

Proof. Let \(X,X_\perp, Y, Y_{\perp}\) be matrices whose columns form a basis for \(F,F^\perp, G, G^\perp\) respectively. Then \((X, X_\perp)\) is unitary. Thus \(M = (X, X_\perp)^\dagger Y\) has orthogonal columns. Therefore we can find \(U_1, U_2, V\) such that \[U^\dagger M V = \begin{pmatrix}U_1^\dagger & 0 \\ 0 & U_2^\dagger\end{pmatrix}\begin{pmatrix}X^\dagger Y \\ X_\perp^\dagger Y\end{pmatrix}V = \begin{pmatrix}U_1^\dagger X^\dagger YV \\ U_2^\dagger X_\perp^\dagger YV\end{pmatrix} = \begin{pmatrix}C \\ S\end{pmatrix}\] Since \(P_F = XX^\dagger\) and \(P_G = YY^\dagger\), putting \(U = (XU_1, X_{\perp} U_2)\), we have \[U^\dagger P_F U = \begin{pmatrix}U_1^\dagger & 0 \\ 0 & U_2^\dagger\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} \begin{pmatrix}U_1^\dagger & 0 \\ 0 & U_2^\dagger\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\] \[U^\dagger P_G U = (U^\dagger Y V)(V^\dagger Y^\dagger U) = \begin{pmatrix}C \\ S\end{pmatrix} \begin{pmatrix}C & S\end{pmatrix} = \begin{pmatrix}C^2 & CS \\ CS & S^2\end{pmatrix}\] Row and column operations corresponding to unitary matrices can then be used to bring this into the desired form. ◻

Proof using group representation theory. Let \(P_F, P_G\) be as before. Put \(R_F = 1 - 2P_F\) and \(R_G = 1-2P_G\) (these are reflections about the given subspaces). One can check that these are unitary operators. Putting \(s = R_F\) and \(r = R_FR_G\), we can see that \(s^2 = (1-2P_F)(1-2P_F) = 1-4P_F + 4P_F^2 = 1\) and \[srs = R_F^2R_G R_F = R_GR_F = r^{-1}\] Thus, the group \(G\) generated by \(R_F, R_G\) is an infinite dihedral group, and \(\langle r \rangle\) is a normal abelian subgroup.

Now we classify the irreducible representations \(V\) of \(G\). Let \(N\) be the group generated by \(r\). There are two cosets in \(G/N\), represented by \(e\) and \(s\). Let \(W \leq V\) be an irreducible representation of \(N\). Then \(W + sW\) is an invariant subspace of \(V\), so \(V = W + sW\). By Shur’s lemma, since \(N\) is Abelian, \(W\) is one-dimensional. This classifies all the irreps, and also gives us a way to construct them. We must have \(r \mapsto e^{i\lambda}\) for some \(\lambda\). Therefore \(V\) is spanned by \(v, sv\), and since \(s^2 = 1\), we have \(s \mapsto \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\) in this basis. Since \(rs = sr^{-1}\), we have \(r = \operatorname{diag}(e^{i\lambda}, e^{-i\lambda})\), where \(\lambda < \pi\) (this is because \(\lambda > \pi\) is clearly isomorphic to a representation with \(\lambda < \pi\)). A basis transformation brings these operators into the form \[\begin{aligned} s &= \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} \\ r &= \begin{pmatrix}\cos(\lambda) & -\sin(\lambda) \\ \sin(\lambda) & \cos(\lambda)\end{pmatrix} \end{aligned}\] Since all unitary representations are completely reducible, we can decompose \(R_F, R_G\) into a block-diagonal form with blocks resembling the above. Then the blocks of the projectors are given by \[P_F = \frac{1-s}{2} = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\] \[P_G = \frac{1-sr}{2} = \begin{pmatrix}\cos^{2}(\lambda/2) & \cos(\lambda/2)\sin(\lambda/2) \\ \cos(\lambda/2)\sin(\lambda/2) & \sin^{2}(\lambda/2)\end{pmatrix}\] Since we can take \(\lambda/2 < \pi/2\), the conclusion is the same. ◻

Proof. Consider \[P_F^\dagger P_G \sim \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} \begin{pmatrix}C^2 & CS \\ CS & S^2\end{pmatrix} = \begin{pmatrix}C^2 & CS \\ 0 & 0\end{pmatrix}\] Then the singular values of the above matrix are the square roots of the eigenvalues of \[\begin{pmatrix}C^2 & CS \\ 0 & 0\end{pmatrix} \begin{pmatrix}C^2 & CS \\ 0 & 0\end{pmatrix}^\dagger = \begin{pmatrix}C^2(C^2+S^2) & 0 \\ 0 & 0\end{pmatrix} = \begin{pmatrix}C^2 & 0 \\ 0 & 0\end{pmatrix}\] Therefore the eigenvalues are just the diagonal entries of \(C\), which are \(\cos(\theta_1), \dots, \cos(\theta_k)\). ◻