Let $Sym^2(V)$ be the set of symmetric matrices of a real $n$-dimensional vector space $V$. Given an element $\underline{\lambda}=[\lambda_1,\ldots \lambda_n]\in \mathbb{RP}^n$, where $\lambda_1\leq\lambda_2\leq\ldots\lambda_n$, I say that a subspace $W\subset Sym^2(V)$ is a $\underline{\lambda}$-space if every matrix $A\in W$ has eigenvalues $\lambda_1(A)\leq\ldots \lambda_n(A)$ such that $[\lambda_1(A), \ldots,\lambda_n(A)]=\underline{\lambda}$. Let $R(\underline{\lambda})$ be the maximal dimension of a $\underline{\lambda}$-space. What I am interested about is the following question:

Question-version 1: Compute $R(\underline{\lambda})$. If not possible, find upper bounds on $R(\underline{\lambda})$.

Example: if all the $\lambda_i$'s are nonzero, then a $\underline{\lambda}$-space $W$ has the property that any (nonzero) $A\in W$ is nonsingular. The problem of finding the maximal dimension $R_H(n)$ of subspaces of nonsingular $n\times n$ symmetric matrices was studied in "On matrices whose real linear combinations are non-singular", by J. F. Adams, Peter D. Lax and Ralph S. Phillips. It turns out that $R_H(n)=\rho(n/2)+1$, where $\rho(n)$ is the Radon-Hurwitz function $\big($$\rho(n)-1$ is the number of linearly independent vector fields on $S^{n-1}$$\big)$. In particular this says that $$R(\underline{\lambda})\leq \rho(n/2)+1$$ if $\lambda_i\neq 0\quad \forall i$.

Another version of the same question is maybe more geometric: Consider the space $Sym^2(V)$, with the (polar) action of the orthogonal group $O(V)$ given by conjugation: $$P\cdot A:=PAP^t,\quad P\in O(V),\\, A\in Sym^2(V).$$

The orbits of this action consist precisely of those matrices which share the same eigenvalues. If $W\subset Sym^2(V)$ is a $\underline{\lambda}$-space, then the unit sphere $W^1$ of $W$ is contained in a $O(V)$-orbit. So it is equivalent to ask:

Question-version 2: Given a $O(V)$-orbit $\mathcal{O}$, find the maximal dimension of a round sphere contained in $\mathcal{O}$

On the one hand it feels that these spaces should have "good upper bounds", meaning it looks hard to come up with examples. Even finding 1-dimensional examples is a bit less trivial than (I) expected. In fact, a matrix $A$ spans a 1-dimensional $\underline{\lambda}$-space, for some $\underline{\lambda}$, if and only if the eigenvalues of $A$ are symmetric around $0$, i.e. if $\lambda$ is an eigenvalue then $-\lambda$ is an eigenvalue as well, with the same multiplicity. On the other hand there are geometric situation (where unfortunately it is hard to compute things directly) that prove the existence of high dimensional $\underline{\lambda}$-subspaces, for very specific $\underline{\lambda}$. Very likely the choice of $\underline{\lambda}$ must be special, and the elements of a $W$ should have special relationships among each other. This leads to my last (very vague) question:

Question 2: Given a n integer $k$, and supposing $R(\underline{\lambda})\geq k$, what can I say about $\underline{\lambda}$, and $W$ being a $\underline{\lambda}$-space?

I am pretty much looking for anything that could help: references, observations for special cases, or why not? complete answers to my questions. :)

Many thanks in advance!

EDIT Even though it doesn't feel too illuminating to me, here is a family of examples (all the ones I know, essentially).

Example 2: Let $0< a_1\leq a_2\ldots \leq a_r$ be real numbers, and consider an embedding of $S:\mathbb{R}^n\to Sym^2(\mathbb{R}^{r(n+1)})$, $\underline{x}\mapsto S_{\underline{x}}$, as follows: if $\underline{w}=(\underline{w}_1,t_1,\underline{w}_2,t_2,\ldots \underline{w}_r,t_r)\in \mathbb{R}^{r(n+1)}$, where $\underline{w}_i\in \mathbb{R}^n$ and $t_i\in \mathbb{R}$, then $$ S_{\underline{x}}=\big(a_1t_1\underline{x}, a_1 \langle \underline{x},\underline{w}_1\rangle, \ldots, a_rt_r\underline{x}, a_r\langle\underline{x},\underline{w}_r\rangle\big),$$ where $\langle,\rangle$ is the canonical scalar product in $\mathbb{R}^n$. One can check that the image of this space is a $\underline{\lambda}$-space, where $$\underline{\lambda}=[-a_r,-a_{r-1},\ldots, -a_1,0,0,\ldots,0,0,a_1,a_2,\ldots, a_r].$$ The number of zeroes in $\underline{\lambda}$, is $r(n-1)$.

This example is a slight generalization of a family of examples arising from a geometric situation. Explicitly, I have a submanifold $M$, of a sphere, a point $p\in M$, and a linear subspace $W$ of the normal space $\nu_pM$ such that in all directions of $W$ the focal points of $M$ arise at the same distances. This means that all the shape operators $S_x$ (which are symmetric endomorphisms of $T_pM$) have the same eigenvalues, so they generate a $\underline{\lambda}$-space.

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    $\begingroup$ These are nice question. Do you have any examples where dimension of $R(\underline{\lambda})$ is $>2$? Trivial example of dimension 2 comes from trace-free symmetric $2\times 2$ matrices. As usual in these kind of questions you probably would need some tools beyond linear algebra; algebraic geometry could be one of these, cf. old work of Eisenbud and Harris on linear subspaces in the space of singular matrices. $\endgroup$ – Misha Mar 15 '12 at 13:19
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    $\begingroup$ There's something I don't understand in your discussion. The orbits of the orthogonal group acting by conjugation on (traceless) symmetric matrices form an isoparametric foliation. In particular, the principal orbits are isoparametric and hence foliated by spheres which are umbilic in the ambient. In fact in this case the spheres are circles. Some of them are centered at the origin (do you want that in your question - version 2?). The situation for the focal manifolds is similar. In other words, the geometry of those orbits is very well understood. What am I missing? $\endgroup$ – Claudio Gorodski Mar 16 '12 at 16:53
  • $\begingroup$ Claudio: Marco is asking for the maximal dimension of round spheres contained in the orbits. Your observation about foliation by circles is that this dimension is $\ge 1$. $\endgroup$ – Misha Mar 16 '12 at 18:03
  • $\begingroup$ Claudio: I totally see what you are saying, and you're right! Let me see if I got this straight: if I prescribe eigenvalues (in my case some coincide) i take the correponding orbit (singular, in my case). Now, this is isoparametric, and I know that some great circles in this orbit correspond to the integral manifolds of eigendistributions of some shape operator. Also, I guess that all great circles arise in this way? That would help a lot... Thanks! If it were an answer I would have probably accepted that already... :) $\endgroup$ – CuriousUser Mar 16 '12 at 20:06
  • $\begingroup$ Marco: Principal orbits are isoparametric, so at each point the family of curvature circles is the same irrespective of the normal direction. On the other hand, the focal orbits, the ones you are interested, do not have flat normal bundle. In this case at each point the family of curvature circles depends on the normal direction, but I think in general you do not get circles in all directions. It has to do with the way curvature circles rotate along other curvature circles in the principal orbits. Of course all this can be explicitly computed without too much effort (I think). $\endgroup$ – Claudio Gorodski Mar 21 '12 at 1:57

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