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.

  1. 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.

  2. 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.

  3. 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?


  1. 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 $A_n\to A$.
  2. No. $\operatorname{diag}(-1,1)$ is the simplest counterexample.
  3. 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.

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

  • $\begingroup$ Why is the first part of #1 true though? I can't think of a proof for it. $\endgroup$ – Leo Spencer Oct 30 '13 at 23:11
  • 1
    $\begingroup$ @Leo, here is a book to start. books.google.com/… $\endgroup$ – Will Jagy Oct 30 '13 at 23:17

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