be subspaces of $\mathbb{R}^n$ spanned by the column vectors of $B$ and $B_k$ respectively. Then for any subspace $\mathbf{A}$ of $\mathbb{R}^n$, the principal angles between the respective subspaces satisfy $$\theta_i(\mathbf{A},\mathbf{B})\le \theta_i(\mathbf{A},\mathbf{B}_k)$$ for $i = 1,\dots,\min(\dim \mathbf{A}, \dim \mathbf{B}_k)$.

By appropriate orthogonalization if necessary, we may assume that $B$ and its submatrix $B_k$ are orthonormal bases of $\mathbf{B}$ and $\mathbf{B}_k$. Let $A$ be an orthonormal basis of $\mathbf{A}$. Then $\sigma_i(A^\mathsf{T}B)$ and $\sigma_i(A^\mathsf{T}B_k)$ take values in $[0,1]$. Since $\theta_i(\mathbf{A},\mathbf{B})= \cos^{-1}(\sigma_i\bigl(A^\mathsf{T}B)\bigr)$ and $\cos^{-1}$ is monotone decreasing in $[0,1]$, the result follows from $\sigma_i(A^\mathsf{T}B)\ge \sigma_i(A^\mathsf{T}B_k)$, by Proposition \[prop:compare\] applied to the submatrix $A^\mathsf{T}B_k$ of $A^\mathsf{T}B$.

\[lem:ineq2\] Let $\mathbf{A}$, $\mathbf{B}$ be as in Lemma \[lem:ineq1\]. Then $\delta(\mathbf{A}, \mathbf{B})\le d_{\operatorname{Gr}(k,n)}\bigl(\mathbf{A},\Omega_{-}(\mathbf{B})\bigr)$.

Let $\mathbf{Y} \in \Omega_{-}(\mathbf{B})$. Then $\mathbf{Y}$ is a $k$-dimensional subspace contained in $\mathbf{B}$ and in the notation of Corollary \[cor:compare\], we may write $\mathbf{Y} = \mathbf{B}_{k}$. By the same corollary we get $\theta_i(\mathbf{A},\mathbf{B})\le \theta_i(\mathbf{A},\mathbf{Y})$ for $i = 1, \dots, k$. Hence $$\label{eq:ineq2}
\delta(\mathbf{A},\mathbf{B})=\Bigl(\sum\nolimits_{i=1}^{k}\theta_i(\mathbf{A},\mathbf{B})^2\Bigr)^{1/2} 
\le \Bigl(\sum\nolimits_{i=1}^{k}\theta_i(\mathbf{A},\mathbf{Y})^2\Bigr)^{1/2}=d_{\operatorname{Gr}(k,n)}(\mathbf{A},\mathbf{Y}).$$ The desired inequality follows since this holds for arbitrary $\mathbf{Y} \in \Omega_{-}(\mathbf{B})$.

Recall that Grassmannians satisfy an isomorphism $$\operatorname{Gr}(k,n)\cong \operatorname{Gr}(n-k,n)$$ that takes a $k$-plane $\mathbf{Y}$ to the $(n-k)$-plane $\mathbf{Y}^{\perp}$ of linear forms vanishing on $\mathbf{Y}$. It is easy to see that 