Let $H$ be a finite-dimensional complex Hilbert space and denote by $d_{\textrm{Haus}}$ the Hausdorff distance between linear subspaces of $H$ i.e., $d_{\textrm{Haus}}(V,W)$ is the usual Hausdorff distance between $V \cap S$ and $W \cap S$, where $S \subseteq H$ is the unit sphere. Consider two linear subspaces $V_1, V_2 \subseteq H$ and denote by $V_1^\perp, V_2^\perp$ their respective orthogonal complements.
How are $d_{\textrm{Haus}}(V_1^\perp, V_2^\perp)$ and $d_{\textrm{Haus}}(V_1, V_2)$ related? Are they equal?
Thank you for your help.
Yes, they are equal. This even works for closed subspaces of infinite-dimensional Hilbert spaces.
Given any $x_1 \in B_{V_1}$ $$d(x_1, B_{V_2}) = \|x_1 - P_{B_{V_2}}x_1\| = \|x_1 - P_{V_2}x_1\| = \|(I - P_{V_2}) x_1\| = \|P_{V_2^\perp} x_1\|,$$ where the second equality holds because projections onto closed, non-empty convex sets are single-valued and non-expansive, and $\|x_1\| \le 1$. To find the Hausdorff distance, we require $$\sup_{x_1 \in B_{V_1}}\|P_{V_2^\perp} x_1\| = \sup_{x \in B_H}\|P_{V_2^\perp} P_{V_1}x\| = \|P_{V_2^\perp} P_{V_1}\|,$$ once again using the non-expansiveness of $P_{V_1}$, and the fact that $P_{V_1} x_1 = x_1$ for all $x_1 \in V_1$. This gives us a characterisation of the Hausdorff distance: $$d_{\text{Haus}}(V_1, V_2) = \max\{\|P_{V_2^\perp} P_{V_1}\|, \|P_{V_1^\perp} P_{V_2}\|\}.$$ Now, since orthogonal projections are self-adjoint linear maps, and $\|A^*\| = \|A\|$ for all bounded operators $A$ under the operator norm, we have $$\|P_{V_2^\perp} P_{V_1}\| = \|(P_{V_2^\perp} P_{V_1})^*\| = \|P_{V_1}^* P_{V_2^\perp}^*\| = \|P_{V_1} P_{V_2^\perp}\|,$$ thus $$d_{\text{Haus}}(V_1, V_2) = \max\{\|P_{V_1} P_{V_2^\perp}\|, \|P_{V_2} P_{V_1^\perp}\|\} = d_{\text{Haus}}(V_1^\perp, V_2^\perp).$$