Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was trying to read Edelman et al.'s 1998 paper "The Geometry of Algorithms with Orthogonality Constraints" and, since I don't have any differential geometry or much linear algebra background, I am stuck in a few places.

This is regarding Section 2.2.1, i.e the tangent and normal spaces of the Stiefel manifold. Here is the excerpt: $\newcommand{\sym}{{\rm sym}}\newcommand{\skew}{{\rm skew}}$

Let $Z$ be any n-by-p matrix. Let $\sym(A)$ denote $(A + A^T)/2$ and $\skew(A)$ denote $(A - A^T)/2$, then at $Y$ , $\pi_N(Z) = Y\sym(Y^TZ)$ defines a projection of Z onto the normal space. Similarly at $Y$, $\pi_T(Z) = Y\skew(Y^TZ) + (I - YY^T)Z$.

I tried deriving $\pi_N(Z)$ as $\pi_N(Z) = {\rm tr}(N^TZ)$; $N = {YS}$ where $S$ is any p-by-p symmetric matrix but couldn't arrive at the result. But using that result I could easily derive $\pi_T(Z)$ because $Z = \pi_T(Z) + \pi_N(Z)$.

Next, they go on to say that tangent directions $\Delta$ at $Y$ then have the general form: $\Delta = YA + Y_{\perp}B$ where $A$ is p-by-p skew symmetric , $B$ is any (n-p)-by-p matrix and $Y_{\perp}$ is any n-by-(n-p) matrix such that $YY^T + Y_{\perp}{Y_{\perp}}^T = 1$.

I couldn't derive the general form of the tangents either. Can someone please help me out on this?


share|cite|improve this question
Minor nitpick: typo in title – Yemon Choi Aug 27 '13 at 5:17
@YemonChoi: why not just fix it, rather than comment? – Stefan Kohl Aug 30 '13 at 12:47
@StefanKohl Usually I prefer to give the OP the chance to do it, as I think MO editing should be done infrequently and with a light touch. However, since we are opening the gates, M-x editmania – Yemon Choi Aug 30 '13 at 20:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.