this isomorphism is an isometry. Using this isometric isomorphism, together with Lemma \[lem:ineq1\] and Lemma \[lem:ineq2\], we can immediately deduce that $$\delta(\mathbf{A}, \mathbf{B}) \le d_{\operatorname{Gr}(k,n)}\bigl(\mathbf{A},\Omega_{-}\bigl(\mathbf{B})\bigr)=d_{\operatorname{Gr}(n-k,n)}\bigl(\mathbf{A^{\perp}},\Omega_{+}\bigl(\mathbf{B}^{\perp})\bigr)\le \delta(\mathbf{A}^{\perp},\mathbf{B}^{\perp}).$$ On the other hand, by results in [@Knyazev], we have $\delta(\mathbf{A}, \mathbf{B}) = \delta(\mathbf{A}^{\perp},\mathbf{B}^{\perp})$ and hence $$\delta(\mathbf{A}, \mathbf{B}) = d_{\operatorname{Gr}(k,n)}\bigl(\mathbf{A},\Omega_{-}(\mathbf{B})\bigr).$$ Similarly we can obtain $$\delta(\mathbf{A}, \mathbf{B}) = d_{\operatorname{Gr}(l,n)}\bigl(\mathbf{B},\Omega_{+}(\mathbf{A})\bigr).$$ Hence we have the required equalities and in Theorem \[thm1\]. Property  is obvious from and Property  follows from Lemma \[lem:infty\].

The proof of Lemma \[lem:ineq1\] provides a simple way to find a point $\mathbf{X} \in \Omega_{+}(\mathbf{A})$ that realizes the distance $d_{\operatorname{Gr}(l,n)}\bigl(\mathbf{B},\Omega_{+}(\mathbf{A})\bigr)=\delta(\mathbf{A},\mathbf{B})$. Similarly we may explicitly determine a point $\mathbf{Y} \in \Omega_{-}(\mathbf{B})$ that realizes the distance $d_{\operatorname{Gr}(k,n)}\bigl(\mathbf{A},\Omega_{-}(\mathbf{B})\bigr)=\delta(\mathbf{A},\mathbf{B})$. One might wonder whether or not Theorem \[thm1\] still holds if we replace $d_{\operatorname{Gr}(k,n)}$ by other distance functions described in Table \[tab:distances\]. The answer is yes.

\[thm:othermetrics\] Let $k \le l \le n$. Let $\mathbf{A}\in \operatorname{Gr}(k,n)$ and $\mathbf{B}\in\operatorname{Gr}(l,n)$. Then $$d^*_{\operatorname{Gr}(k,n)}\bigl(\mathbf{A}, \Omega_{-}(\mathbf{B})\bigr) =  d^*_{\operatorname{Gr}(l,n)}\bigl(\mathbf{B}, \Omega_{+}(\mathbf{A})\bigr),$$ for $* = \alpha, \beta, \kappa, \phi, \mu, \rho, \pi, \sigma$. Their common value $\delta^*(\mathbf{A},\mathbf{B})$ is given by: $$\begin{aligned}
\delta^{\alpha}(\mathbf{A}, \mathbf{B}) &=  \theta_k, &
\delta^{\beta}(\mathbf{A},\mathbf{B}) &= \Bigl(1 - \prod\nolimits_{i=1}^k\cos^2\theta_i\Bigr)^{1/2},\\
\delta^{\kappa}(\mathbf{A}, \mathbf{B}) &= \Bigl(\sum\nolimits_{i=1}^k\sin^2\theta_i\Bigr)^{1/2}, &
\delta^{\phi}(\mathbf{A}, \mathbf{B}) &= \cos^{-1}\bigl(\prod\nolimits_{i=1}^k\cos \theta_i\Bigr),\\
\delta^{\mu}(\mathbf{A}, \mathbf{B}) &= \Bigl( \log \prod\nolimits_{i=1}^k\frac{1}{\cos^2 \theta_i}\Bigr)^{1/2}, 