# Set of symmetric positive semidefinite matrices is closed

I am self-studying Boyd & Vandenberghe's Convex Optimization.

Example 2.15 (page 43) states that the symmetric positive semi-definite cone $$S^n_+$$ is a proper cone. This necessitates, amongst other things, that it is closed.

I am not sure how to show that $$S^n_+$$ is closed, particularly because this set consists of matrices, which I am less comfortable working with.

The most relevant question I have found that may have some relation to this one is here; I am not sure how to act on the answer of this question for I am not sure of whether the functions $$f_1$$ and $$f_2$$ as defined in the answer are relevant to my task.

$$S_+^n$$ closed follows quite elementarily from definition, rather than by using topological properties of the eigenvalues of a matrix.

Let $$x\in\Bbb R^n$$ and consider the following linear maps:

• $$G:\Bbb R^{n\times n}\to \Bbb R^{n\times n}$$, $$G(A)=A-A^t$$.

• $$q_x:\Bbb R^{n\times n}\to \Bbb R$$, $$q_x(A)=x^tAx$$.

Since all these maps are continuous, $$S^n_+:=\ker G\cap \bigcap_{x\in\Bbb R^n} q_x^{-1}[0,\infty)$$ must be closed, because it's intersection of closed sets. As an additional observation, this is also an intersection of preimages of convex cones by linear maps, and thus a convex cone.

The space $$\mathbf{R}^{n \times n}$$ is a $$(n^2)$$-dimensional real vector space, and the space $$\mathbf{S}^n$$ of symmetric matrices is a linear subspace (this is easy to check). The map $$\lambda_{\min} : \mathbf{S}^n \to \mathbf{R}$$, given, for example, by $$\lambda_{\min}(X) = \min_{\|v\| = 1}v^TXv$$ is continuous (with respect to the relative topology on $$\mathbf{S}^n$$). Now note that $$\mathbf{S}^n_+ = \{X \in \mathbf{S}^n: \lambda_{\min}(X) \geq 0\} = \lambda_{\min}^{-1}([0, \infty)),$$ which is the continuous preimage of a closed set, thus closed.

• How to prove $\lambda_{\min}(X)$ is continuous, can you provide some references, thanks in advance!
– Kim
Commented Jan 29 at 13:43
• Start by noticing that $v^T(X - X') v \leq \|X - X'\| \|v\|^2$. This implies that $\lambda_{\rm min}$ is Lipschitz continuous. Commented Jan 30 at 23:01

First, observe that $$S^n_+$$ is self-dual (e.g. see Boyd's example 2.24).

Then note that the dual cone, $$K^*$$ is closed and convex (since, by definition, the dual cone is the intersection of a set of closed halfspaces; and since the intersection of closed sets is closed, and since the intersection of any number of halfspaces is convex).

So, since $$S^n_+ = (S^n_+)^*$$, and since the latter is closed and convex, we know that $$S^n_+$$ is closed and convex! (and self-dual!)