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$\newcommand{\GLp}{\operatorname{GL}_n^+}$ $\newcommand{\SO}{\operatorname{SO}_n}$ 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.

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  • $\begingroup$ My approach would be to try to show that the complement is an algebraic set of codimension at least 2. That would give the result. $\endgroup$ – user98602 Jan 1 at 17:25
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    $\begingroup$ Unfortunately your new last line is false. The complement contains eg the hyperplane $(x_1, x_1, x_3, \cdots, x_n)$, which separates your set into two regions. $\endgroup$ – user98602 Jan 1 at 17:28
  • $\begingroup$ I changed my mind as I started writing up the heuristic. I think connectedness is probably true, along the lines of the first comment. In the setting of eigenvalues, my intuition is that if you just want the dimension count, you may as well just study the set for which $A$ has two-dimensional kernel, and then take the product with $\Bbb (-\epsilon, \epsilon)$ to model 2-dimensional eigenspaces near zero. Then the space of matrices with $\text{ker}(A)$ of dimension 2 is $(n-2)^2$, in particular codimension $4n -4$. So if $4n \geq 5$ - that is, $n > 1$ - I expect codimension larger than 2. $\endgroup$ – user98602 Jan 1 at 17:38
  • $\begingroup$ Instead of posting it as an answer I will say that my comment above is essentially a proof in the case of eigenvalues. I don't know anything about singular values so won't post it. Since you deleted your comments from earlier, I should ask if I said anything wrong up above? $\endgroup$ – user98602 Jan 1 at 18:12
  • $\begingroup$ No, no. Everything you said was great. I deleted since there were too many comments:) By the way, I am not entirely sure I understood your heuristic: What do you mean by "multiplying by $(-\epsilon,\epsilon)$"? I imagine matrices with two-dim kernel as a space of dimension $n(n-2)$; I think of an $n \times n$ matrix with two zero columns. But somehow you got to $(n-2)^2$ (I guess by accounting somehow nearby eigenspaces...). Is there any chance you could elaborate on that? $\endgroup$ – Asaf Shachar Jan 1 at 18:33
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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 $GL_n^+$.

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  • $\begingroup$ Thanks; so let me make sure that I understand you correctly: If $A=U_1\Sigma_1V_1^T$ and $A_2=U_2\Sigma_2V_2^T$ where $\Sigma_1,\Sigma_2$ are strictly decreasing, then since $SO$ is connected, there is a path from $A_1$ to $U_2\Sigma_1V_2^T$, and now we pass from $\Sigma_1$ to $\Sigma_2$ via convex combinations. (So we get from $U_2\Sigma_1V_2^T$ to $U_2\Sigma_2V_2^T=A_2$). Is that it? $\endgroup$ – Asaf Shachar Jan 1 at 19:40
  • $\begingroup$ @AsafShachar Yes. Suppose $U_i$s and $V_i$s have been chosen as members of $SO_n$ and the diagonals of $\Sigma_1,\Sigma_2$ are strictly decreasing tuples. Then there exist skew-symmetric matrices $K_1,K_2,K_3,K_4$ such that $U_1=e^{K_1},\ V_1=e^{K_2},\ U_2=e^{K_3}$ and $V_2=e^{K_4}$. Now $f(t)=e^{(1-t)K_1+tK_3}[(1-t)\Sigma_1+t\Sigma_2]e^{-(1-t)K_2-tK_4}$ is a path joining $A$ and $B$ in $X$. $\endgroup$ – user1551 Jan 1 at 19:46
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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 $n \geq 1$.

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  • $\begingroup$ Thanks, that is a nice explanation. Could you say a bit more about the transversality argument? that is, assuming that the set of matrices with non-distinct eigenvalues are stratified by the different submanifolds corresponding to the different partitions of $n$, and that they are all smooth submanifolds of codimension at least 2, then how does the connectedness claim follow by transversality? $\endgroup$ – Asaf Shachar Jan 2 at 9:34
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    $\begingroup$ @AsafShachar: Let $S$ be a subspace of some manifold $M$ so that $S$ is a finite union of submanifolds of codimension larger than 1. If $I \to M$ is a curve whose endpoints miss $S$, then the transversality theorem says that for any one of those submanifolds in $S$, we may perturb our map away from the boundary by an arbitrarily small amount to be transverse to that submanifold - hence, disjoint from it. Now, maps transverse to a given submanifold form an open set. So doing this iteratively, we construct a map transverse to all of the submanifolds in $S$, hence disjoint from $S$. $\endgroup$ – user98602 Jan 2 at 9:45

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