# Is the space of $n\times n$ real symmetric matrices with strictly positive determinant connected within the vector space of $n\times n$ real matrices?

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.

• 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. May 12, 2022 at 14:19
• 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). May 12, 2022 at 14:27
• @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. May 12, 2022 at 14:36
• @Hvjurthuk See this post regarding the first part of my suggested proof May 12, 2022 at 14:39
• It does not change anything. Indeed, such a space is still not connected for the same reason. May 12, 2022 at 14:43

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.