Questions on positive definite matrices First, in this discussion, I am only considering real matrices.  Second, I have a few questions I am ruminating on related to symmetric matrices.  Some of these questions I need someone to say my logic is correct while others I need you to help provide the logic.


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*I am curious as to what the closure of the set of positive definite matrices is.  I know that this set is an open cone.  Just thinking about it seems to point to the closure being the positive semi-definite matrices.  However, I cannot think of a way to prove this.  Also, is the closure in the set of symmetric matrices, the same as the closure in the set of all matrices.  I believe this to be the same as any sequence of symmetric matrices should be a symmetric matrix.

*Also, is every symmetric matrix, either positive definite, negative definite or in the boundary of both?  Again this seems somewhat logical but I don't know why.

*Finally, if the closure of the positive definite matrices is the positive semi-definite ones, do all the matrices in the boundary have determinant zero?  This seems true as I believe these matrices have at least one eigenvalue as $0$ implying $0=det(A-\lambda I)= det (A-0I) =det (A)$.  If I was wrong about the closure is there any way to characterize the matrices in the boundary?
Thanks!
 A: 1 and 3 are correct. Remember that the symmetric matrices make a vector subspace. 
What you are missing is the indefinite quadratic forms, such as $x^2 - y^2.$ This corresponds to a diagonal matrix with diagonal entries $(1,-1).$ It is not in the boundary of anything you have mentioned. 
As the dimension $n$ goes up in symmetric $n$ by $n$ matrices, you get more and more separate connected components, according to signature: http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
A: *

*Yes. Suppose $\{A_n\}$ is a family of PD matrices and $A_n\to A$. Then $0=A_n-A_n^\ast\to A-A^\ast$ (hence $A$ is Hermitian) and for any vector $x$, we have $x^\ast A_nx>0$ and $x^\ast A_nx\to x^\ast Ax$ (hence $x^\ast Ax\ge0$). It follows that $A$ is PSD. Conversely, if $A$ is PSD, then $A_n=A+\frac1n I$ is PD and $A_n\to A$.

*No. $\operatorname{diag}(-1,1)$ is the simplest counterexample.

*Yes. The set of PD matrices is open (this can be proved by considering $x^\ast(A+\delta A)x$). So, every matrix on the boundary of PD matrices is a PSD matrix with at least one zero eigenvalue and therfore it has zero determinant.

