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I want to make clear that I am aware of the connectedness in the case of general real matrices. But here I ask about the subspace of symmetric ones.

If it is not the case, which are the connected components of such topological space? If it is the case, what would be the path on such space connecting say a signature matrix $J$ with positive determinant with the identity $I$? That is, give a nontrivial path of symmetric matrices with positive determinant from some signature matrix $J\neq I$ with $\det(J)=1>0$ to the identity $I$.

Remember that a signature matrix is a diagonal matrix whose diagonal entries belong to $\{-1,1\}$. Note also that, as this matrix in my question has to have positive determinant $-1$ appears an even number of times in such diagonal.

Hvjurthuk
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  • Does this answer your question? Prove that the set of $n$-by-$n$ real matrices with positive determinant is connected – Jean Marie May 12 '22 at 14:16
  • What is a signatura matrix ? – Jean Marie May 12 '22 at 14:17
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    Dear Jean, no it does not answer my question because I need the matrices to be symmetric. Sorry, I meant a signaturE matrix: a diagonal one having only 1's and -1's in the diagonal. – Hvjurthuk May 12 '22 at 14:19
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    As you have correctly noted, signature plays an important role here. In the $2 \times 2$ case, there are two connected components: one consisting of the elements with purely positive signature (or equivalently given the constraints, a positive upper-left entry) and those with a purely negative signature (or equivalently a negative upper-left entry). – Ben Grossmann May 12 '22 at 14:27
  • Dear Ben, okay, then do we actually have that each signature with positive determinant defines a connected component of this set in general? Do you have a reference? That would be also helpful. – Hvjurthuk May 12 '22 at 14:30
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    @Hvjurthuk Yes. One can form a path connecting two elements of the same signature by using the fact that the orthogonal matrices with determinant one form a path-connected set. The trickier part is showing that two matrices of distinct signature do not lie in the same connected component; one approach is to use the continuous dependence of eigenvalues on matrix entries. I don't have a reference offhand. – Ben Grossmann May 12 '22 at 14:36
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    @Hvjurthuk See this post regarding the first part of my suggested proof – Ben Grossmann May 12 '22 at 14:39
  • @BenGrossmann, does anything change if I substitute symmetric by Hermitian and consider complex matrices instead of real ones? Is such space still not connected? – Hvjurthuk May 12 '22 at 14:42
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    It does not change anything. Indeed, such a space is still not connected for the same reason. – Ben Grossmann May 12 '22 at 14:43
  • @BenGrossmann Ok, thanks for your insights! – Hvjurthuk May 12 '22 at 14:44
  • Just out of curiosity, why is this interesting? What can we do with this information? – Filippo May 12 '22 at 21:20

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Great question!

This answer is inspired by @Ben Grossman's comment. We rely on the fact that eigenvalues are continuous functions of the entries of the matrix. (see this answer eigenvalues-are-continuous)

Call the space of symmetric matrices $S^{n\times n}$. The space of symmetric matrices with positive determinant $S^{n\times n}_+$. Then consider a symmetric matrix $M$ with more than 1 (at least 2) negative eigenvalues, and the identity matrix $I$. Consider a path connecting them $$\gamma:[0,1]\to S^{n\times n},t\mapsto A(t)\\ \gamma(0)=M,\gamma(1)=I $$. Such path is easy to construct $\gamma(t)=M+t(I-M)$

Then consider the eigenvalues of $A(t)$ along the path. There exist a continuous function $f_i:\mathbb R\to\mathbb R,t\mapsto \lambda_i(A(t))$ (real symmetric matrices have real eigenvalues), $i$ denotes the $i$th largest eigenvalue.

Then find the negative eigenvalues of $M$ let the eigenvalue be $\lambda_1$ then corresponding function $f_1(t)$ has $f_1(0)=\lambda_1<0,f_1(1)=1$ Then with the property of continuous function, there exist $t_*$ that $f(t_*)=0$, then $\det A(t_*)=0$, $A(t_*)\not\in S^{n\times n}_+$.

Thus for any path between $M$ and $I$ in the space of real symmstric matrices, the path cannot stay in $S^{n\times n}_+$, then we know $S^{n\times n}_+$ is not path connected.

Binxu Wang 王彬旭
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