Cfrgtkky

$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
Consider the set
$$ X={ A in GLp ,,|,, text{ all the singular values of } A text{ are distinct } },$$
where $GLp$ is the group of real $ntimes n$ 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 $A=USigma V^T in GLp$ 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.

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 $Ain X$ 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 $GL_n^+$.

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 $n geq 1$.