I want to minimize the distance between two orthonormal k-frames in $R^{k+1}$.
The first k-frame is $\{e_1,...,e_k\}$ where $e_i = (0,...,0,1,0,,,0)$ denote i'th unit vectors in $R^{k+1}$. The second k-frame is a variable for this optimization problem, with restriction that for given vector $v \in R^{k+1}$, this k-frame $\{u_1,...,u_k\}$ lies in subspace orthogonal to $v$.
The 'distance' here means canonical distance in Stiefel manifold, which is simply the square root of the sum of squares of Euclidean distance between each elements of k-frames. That is, the objective function is $\sqrt{\sum_{i=1}^{k} \sum_{j=1}^{k+1}(u_{ij}-e_{ij})^2}$.
I wonder if there is an explicit solution to the optimization problem above. If not, is there a way to solve this problem with computing power?
Let $e=(e_1,\ldots,e_n)$ be the canonical basis of the Euclidean space $R^n.$ Let $E=R^k\times \{0\}\subset R^n$, let $a$ an be an orthogonal endomorphism of $R^n$, the space $F=a(E)$ and finally $u_j= a(e_j)$ for $j=1,\dots,n.$ Therefore $(u_1,\ldots,u_k )$ is an orthonormal basis of $F.$ Consider now an orthogonal endomorphism $b$ of $F.$ We extend it to $R^n$ by defining $b(x)=x$ when $x\in F^{\perp}.$ Since $(b(u_1),\ldots,b(u_k))$ is another orthogonal basis of $F$ we are interested in minimizing $$ b\mapsto \sum _{i=1}^k\|e_i-b(u_i)\|^2=2k-2\sum _{i=1}^k\langle e_i,b(u_i)\rangle$$ or maximizing $$b\mapsto \sum _{i=1}^k\langle e_i,b(u_i)\rangle$$
Now assume that we know the column vectors $U_1,\ldots,U_k$ which are the components in the basis $e$ of the vectors $u_1,\ldots,u_k$ and denote by $P$ the $(k,k)$ matrix made with the first $k$ rows of the $(n,k)$ matrix $[U_1,\ldots,U_k].$ Furthermore let $B\in O(k)$ be the orthogonal matrix which is the representative matrix of $b$ in the basis $u=(u_1,\ldots,u_k)$ of $F.$ We claim that $$\sum _{i=1}^k\langle e_i,b(u_i)\rangle=\mathrm{trace}(PB).\ \ (*)$$ To see this denote consider the representative matrix $[a]_e^e$ in the base $e$, this is also the matrix $\mathrm{id_{R_n}}$ of the identity map in $R^n$ with starting basis $u$ and landing basis $e$, in other terms the matrix whose columns are the components of the $u_i$'s expressed in the basis $e$, also called the change basis matrix basis from the old $e$ to the new basis $u$. We write $$ [b]_u^e=[\mathrm{id_{R^n}}]_u^e \times [b]^u_u=\left[\begin{array}{cc}P&P_{12}\\P_{21}&P_2\end{array}\right]\left[\begin{array}{cc}B&0\\0&I_{n-k}\end{array}\right]=\left[\begin{array}{cc}PB&P_{12}\\P_{21}B&P_2\end{array}\right]$$ and we observe that $$[b]_u^e=(\langle e_i,b(u_j)\rangle)_{1\leq i,j\leq n}.$$ This proves (*).
Now we observe that the max of $$B\mapsto \mathrm{trace}(PB)$$ when $B\in O(k)$ is easy to compute. To this end we use the polar decomposition of $P=UDV$ where $D$ is diagonal with non negative coefficients and $U,V\in O(k).$
The key point is the following fact: suppose that $U=V=I_k.$ Then the maximum on $O(k)$ of $B\mapsto \mathrm{trace}(DB)=d_1b_{11}+\cdots+d_kb_{kk}$ is reached when $B=I_k$ since $d_i\geq 0$ and $|b_{ii}|\leq 1. $ Therefore in the general case $$\mathrm{trace}(PB)=\mathrm{trace}(DVBU)$$ is maximum when $VBU=I_k$ or $$\boxed{B=V^{-1} U^{-1}=(UV)^T}$$