# Find the point on the Stiefel Manifold that is closest to a matrix

I don't have much background on high-dimensional geometry, so I dare to ask it.

For a given point in $x\in\mathbb{R}^n$, assume that I want to find the point on the unit sphere that is closest to the point $x$. It's easy -- this can be computed by constructing a line through $x$ and the origin, and find the point where this line intersects with the sphere. I'd like to generalize this to $\mathbf{X}\in\mathbb{R}^{n\times k}$ where the sphere is replaced with a Stiefel manifold $\mathbf{Z}^T\mathbf{Z}=\mathbf{I}_{k\times k}$.

Originally the columns of $\mathbf{X}$ constitute the orthonormal basis of $k$-dimensional subspace of $\mathbb{R}^n$, which means $\mathbf{X}$ is on the Stiefel manifold. But the matrix is perturbed by some random matrix $\mathbf{E}$ such that $\tilde{\mathbf{X}}=\mathbf{X}+\mathbf{E}$. I want to find the closest point of $\tilde{\mathbf{X}}$ that is on the Stiefel manifold. In this case, the distance between two matrices $\mathbf{X}$ and $\mathbf{Y}$ must be measured by comparing their orthonormal projectors $||\mathbf{X}\mathbf{X}^T - \mathbf{Y}\mathbf{Y}^T||_2$ to avoid technical issues with ordering and rotations among the basis vectors.

I think there exists only one point $\mathbf{Y}$ on the Stiefel manifold that is closest to $\tilde{\mathbf{X}}$, but I'm not sure how to find it. One thing I can try is employing Lagrange multipler and try to minimize $||\tilde{\mathbf{X}}\tilde{\mathbf{X}}^T - \mathbf{Y}\mathbf{Y}^T||_2$ subject to $\mathbf{Y}^T\mathbf{Y}=\mathbf{I}$, but I'm not sure whether I have to compute derivative of a spectral norm of a matrix. Is there any 'natural' computation to find it?

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I imagine you're looking for the Cholesky decomposition. See the "Alternatives" part of the Gram-Schmidt Wikipedia page: en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process –  Ryan Budney Mar 28 '11 at 20:52
Take a look at en.wikipedia.org/wiki/Orthogonal_Procrustes_problem –  Federico Poloni Mar 28 '11 at 20:59
oops, sorry, that's the solution for the Frobenius-norm proximity –  Federico Poloni Mar 28 '11 at 21:08
@Ryan Budney. Thank you very much for your quick response! It seems that I can apply the Cholesky decomposition in the least squares problem. –  Federico Magallanez Mar 28 '11 at 23:36

Are you sure you need the distance function $\|XX^T-YY^T\|_2$? If not, the solution to $\min {\|E\|: E \in \mathbb{R}^{m \times n}, (X+E)^T(X+E) = I }$ for the 2-norm and the Frobenius norm is the polar decomposition; see, e.g., section 4 in http://www.maths.man.ac.uk/~nareports/narep161.pdf .
Maybe you do want $\|XX^T-YY^T\|_2$ as distance function because of the equivalence by the orthogonal group. In that case you maybe want to rephrase your problem on the Grassmann manifold on linear subspaces. In fact, the Stiefel manifold with the orthogonal group factored out is exactly the Grassmann manifold. There is a nice paper about this: http://dx.doi.org/10.1137/S0895479895290954 .