Span of positive (semi)definite matrices

Question

In this answer, Ben Grossmann said that $\operatorname{span}\{P:P \succ 0\} = \{A:A = A^T\} =: \mathcal S$. I am not sure, however, why this is true. Also, I guess $\operatorname{span}\{ P:P\succeq 0 \} = \operatorname{span}\{ P:P \succ 0 \}$?

I tried to use symmetric eigenvalue decomposition and also $P=B^{2} = CC^{T}$ but both did not work. So how can we write a symmetric matrix as a linear combination of positive (semidefinite) matrices?

Answer 1

Every linear combination of symmetric positive matrices is a symmetric matrix. Therefore the linear span of positive definite matrices is a subset of all symmetric matrices.

For the reverse inclusion, observe that for any symmetric matrix $A$, the sum $A+cI$ will be positive definite matrix when $c>0$ is sufficiently large. (In particular, $A+cI$ is guaranteed to be positive definite if $c>$ the spectral radius of $A$.) Hence $A=(A+cI)-cI$ is the linear combination of two positive definite matrices $A+cI$ and $cI$, i.e., $A$ lies inside the linear span of all positive definite matrices.

Answer 2

You should try to find a basis of $S_n (\mathbb{R})$ in $S_n^{+} (\mathbb{R})$ from a basis you already know, and then adapt it to $S_n^{++}$. I recall that both of these spaces are convex.

A quicker answer would invoke the continuity of the spectrum, thus if you take $B = \mathbb{B} (I_n, \epsilon) \cap S_n ( \mathbb{R})$ for $\epsilon$ small enough and the canonical norm given by the trace of $A^T A$, you have $B \subset S_n^{++} (\mathbb{R})$.And Span$(B) = S_n (\mathbb{R})$.

Answer 3

Clearly the span of positive semi definite matrices is contained in the space of symmetric matrices. To show the reverse inclusion, given $S$ symmetric, write $S=U\Sigma U^T$. Then write $\Sigma = \Sigma^+-\Sigma^-$ where we apply the $x\mapsto x^+$ and $x\mapsto x^-$ functions to $\Sigma$ entry wise. Thus $S=U(\Sigma^+-\Sigma^-)U^T$, the difference of two positive semi definite matrices.