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The following theorem appears without proof in :

Helmke, Uwe, and John B. Moore. Optimization and dynamical systems. Springer Science & Business Media, 2012.

Let $A$ be a symmetric $n\times n$ real matrix. Define the Stiefel manifold as $St(k,n)=\{X\in \mathbb{R}^{n\times k}|X^TX=I\}$.

Then, we consider the following equation :

$\dot{X}=(I-XX^T)AX$, where $X\in St(k,n)$

It is a matrix ODE which is invariant under right multiplication by $O(k)$. Hence it can be considered as an ODE on the Grassmannian.

Helmke and Moore state that it almost-surely converges to an $A$-invariant subspace spanned by a dominant $k$-dimensional eigenbasis of $A$.

I am looking for proof of this theorem. Does anyone know any suitable references ?

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    $\begingroup$ I think the statement is not quite correct: The equation is compatible with right multiplication by $G\in O(k)$. This means that the best one can hope for is that the limit $X_\infty$ spans a subspace also spanned by $k$ dominant eigenvectors. Or in other words, this should more naturally be considered as an ODE on a Grassmann manifold. $\endgroup$ Commented Mar 13, 2017 at 17:08
  • $\begingroup$ @SebastianGoette Lets consider the vector field on St(k,n) not Grassmanian.Is it true (and obvious) that the columns of every singularity $X$ of the vector field span a k dimensional dominant eigen space? $\endgroup$ Commented Mar 13, 2017 at 20:17
  • $\begingroup$ @mystupid_acct What is a description of the singularities of the vector field. Please see my previous comment. $\endgroup$ Commented Mar 13, 2017 at 20:18
  • $\begingroup$ @SebastianGoette You are correct, i have modified my question. $\endgroup$ Commented Mar 13, 2017 at 22:02
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    $\begingroup$ @AliTaghavi If the columns of $X$ span an invariant subspace of $A$, then $X$ is a fixpoint. This is because $(1-XX^T)$ projects onto the orthogonal complement of the span of the columns of $X$. But not all fixpoints are stable ... $\endgroup$ Commented Mar 13, 2017 at 22:21

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On the Stiefel manifold, consider the function $f(X)=\operatorname{tr}(X^TAX)$. It evolves under the flow of the given vector field as $$\frac d{dt}f=2\operatorname{tr}(X^TA(1-XX^T)AX)\;.$$ Because $1-XX^T$ describes the projection onto the orthogonal complement of the span $V_X$ of the columns of $X$, we have \begin{gather*}\frac d{dt}f\ge 0\;,\\\frac d{dt}f=0\iff \dot X=0\iff\text{$V_X$ is $A$-invariant.}\end{gather*} Hence, $X$ is a generalised gradient field for $f$ (it might in fact be half the actual gradient, but I have not checked).

Now it is easy to see that $f$ induces a Morse-Bott function on the Grassmannian (Morse if the eigenvalues of $A$ are distinct). The only stable fixpoints therefore correspond to the maximum, and the points that don't flow into one of the maxima form a stratified subset of lower dimension.

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  • $\begingroup$ Is it easy to compute the index of each invariant subspace as a singularity of the morse function? $\endgroup$ Commented Mar 14, 2017 at 16:03
  • $\begingroup$ @AliTaghavi Yes, I think so. Take a basis of eigenvectors and see what happens if you modify one of them at a time. $\endgroup$ Commented Mar 14, 2017 at 21:03

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