Is the set of matrices with distinct singular values connected and dense?

Question

$\newcommand{\GLp}{\operatorname{GL}_n^+}$ $\newcommand{\SO}{\operatorname{SO}_n}$ Consider the set $$ X=\{ A \in \GLp \,\,|\,\, \text{ all the singular values of } A \text{ are distinct } \},$$ where $\GLp$ is the group of real $n\times n$ invertible matrices with positive determinant.

Is $X$ connected? (with the subspace topology induced by $\GLp$). Is $X$ dense in $\GLp$?

Since the singular values of $A$ are the eigenvalues of $\sqrt{A^TA}$, I guess this might be connected to the question:

Are matrices in $\GLp$ with distinct eigenvalues dense (in $\GLp$) and connected? I am quite sure they should be dense, but I am not sure about the connectedness.

Edit:

I am quite sure that the density is OK. Indeed, the singular values $\sigma_i$ are distinct if and only if their squares $\sigma_i^2$ are distinct. The $\sigma_i^2$ are the eigenvalues of $A^TA$. Now the argument here with the discriminant of the characteristic polynomial of $A^TA$ works.

Here is an idea regarding the connectedness:

Let $A=U\Sigma V^T \in \GLp$ be with distinct singular values. Since we can always assume that $U,V \in \SO$ and $\SO$ is connected, we can always connect $A$ to $\Sigma$ by "moving on the sides" in $\SO$. (Since we are only changing the orthogonal components, the singular values remain constant along this path).

So, we only need to move between the different $\Sigma$'s. That is, we are left with the following question:

Is the set of vectors in $(\mathbb{R}^{\ge 0})^n$ with different entries connected? I guess there should a be a slick argument deciding this either way.

Answer 1

To answer your second question: no, the subset of $(\mathbb R^{\ge0})^n$ containing vector with distinct coordinates is not necessarily connected. Consider the case where $n=2$. The first quadrant of the $x-y$ plane is divided by the line $x=y$ into two connected components.

But the answer to your first question is yes, because we can perform singular value decomposition on every $A\in X$ so that the singular values are arranged in decreasing order (and the orthogonal matrices have determinants $1$). Since any convex combination of two strictly decreasing tuples is again a strictly decreasing tuple (and, as you said, $SO_n$ is path-connected), $X$ is path-connected. As each decreasing tuple is the limit of a sequence of strictly decreasing tuples, $X$ is dense in $GL_n^+$.

Answer 2

This only answers the easier second question about having distinct eigenvalues, because I don't know anything about singular values.

I claim that the set of matrices for which all but one eigenspaces are 1-dimensional, and the remaining one is 2-dimensional is a smooth submanifold of $GL_n^+$; in fact, one should be able to prove that given any partition of $n$, the set of matrices whose eigenvalues partition $n$ in that way is a smooth submanifold. The proof would be more subtle, though, and I'm not terribly interested in the details. Anyway, the set of matrices with non-generic eigenvalues are stratified by these, and if they are all smooth submanifolds of codimension at least $2$, then the connectedness claim follows by transversality.

Here is my approach; fix a given matrix $A$. There is no harm in assuming that the double eigenvalue is $0$. If I can show you that the set of matrices whose only double eigenvalue is $0$ is a smooth manifold, then the same is true for the matrices whose double eigenvalue is arbitrary: if the former set is written $X$, the latter is $X \times \Bbb R$, the canonical bijection given by $(A, t) \mapsto A + tI$.

So it suffices to verify that the set of matrices with $\dim \text{ker}(A) = 2$ and all other eigenvalues isolated is a smooth manifold; this is an open subset of the set of matrices with $\dim \text{ker}(A) = 2$, and so we just need the result for the latter. This result is a well-known exercise which I will not repeat here; the dimension of $X$ is $(n-2)^2$, and hence the codimemsion of $X \times \Bbb R$ is $4n-3$.

Because we wanted $4n - 3 \geq 2$, we see that this holds for the usual eigenvalue problem whenever $4n \geq 5$; that is, whenever $n \geq 1$.