Let $V$ be an $n$-dimensional real inner product space. Consider the space $V \otimes V$, endowed with the tensor product metric, i.e.
$$ \langle v_1 \otimes v_2 , w_1 \otimes w_2\rangle := \langle v_1, w_1 \rangle_V \cdot \langle v_2 ,w_2 \rangle_V.$$
Define $$S=\{ \sigma \in V \otimes V \, | \, \langle \sigma ,v \otimes v \rangle \ge 0 \, \, \text{ for every $v \in V$} \}$$
Question: Does every $\sigma \in S$ admit a representation in the form of
$\sigma=\sum_i a_i v_i \otimes v_i$, for some $a_i \ge 0$ and $v_i \in V$?
Clearly every such element belong to $S$. Note that $S$ is a convex cone inside $V \otimes V$.
I tried to start with some orthonormal basis $e_i$ for $V$, and write $\sigma=a_{ij}e^i \otimes e^j$ (with the summation convention); testing against $w \otimes w$ for $w=e^i \pm e^j$ I deduced that $a_{ii} \ge 0 $ and $|a_{ij}+a_{ji}| \le a_{ii}+a_{jj} $. I am not sure however whether $\sigma \in S$ implies that $a_{ij}=0$ for $i \neq j$ in general.
Best Answer
No, because $\sigma$ "might be alternating". However each $\sigma$ has such representation up to an arbitrary alternating tensor. Consider the maps $\operatorname{Alt}, \operatorname{Sym} \colon V \otimes V \rightarrow V \otimes V$ given (on elementary tensors and extended linearly) by
$$ \operatorname{Alt}(v \otimes w) = \frac{v \otimes w - w \otimes v}{2}, \,\,\, \operatorname{Sym}(v \otimes w) = \frac{v \otimes w + w \otimes v}{2}. $$
One can verify that $\operatorname{Alt},\operatorname{Sym}$ are orthogonal complementary projections onto the space of alternating and symmetric tensors respectively.
Now, given $\sigma \in S$, consider the symmetric bilinear form $g_{\sigma}(v,w) := \left< \sigma, \operatorname{Sym}(v\otimes w) \right>$. Note that $$ g_{\sigma}(v,w) = \left< \operatorname{Alt}(\sigma) + \operatorname{Sym}(\sigma), \operatorname{Sym}(v \otimes w) \right> = \left< \operatorname{Sym}(\sigma), \operatorname{Sym}(v \otimes w) \right> = g_{\operatorname{Sym}(\sigma)}(v,w) $$ so that $g_\sigma$ depends only on the symmetric part of $\sigma$. By assumption, we have $$g_{\sigma}(v,v) = \left< \sigma, \operatorname{Sym}(v \otimes v) \right> = \left< \sigma, v \otimes v \right> \geq 0 $$ so $g_{\sigma}$ is positive semidefinite. Hence, we can find an orthonormal basis $(e_i)_{i=1}^n$ for $V$ such that $g_{\sigma}(e_i,e_j) = 0$ if $i \neq j$ and $g_{\sigma}(e_i,e_i) = a_i \geq 0$. If we write $\sigma$ with respect to this basis as $\sigma = \sigma^{kl} e_k \otimes e_l$ we get
$$ g_{\sigma}(e_i, e_j) = \left< \sigma^{kl} e_k \otimes e_l, \frac{e_i \otimes e_j + e_j \otimes e_i}{2} \right> = \frac{\sigma^{ij} + \sigma^{ji}}{2} = 0, \,\, i \neq j, \\ g_{\sigma}(e_i,e_i) = \left< \sigma^{kl} e_k \otimes e_l, e_i \otimes e_i \right> = \sigma^{ii} = a_i \geq 0.$$
Hence, we can write $\sigma$ as
$$ \sigma = \operatorname{Alt}(\sigma) + \sum_{i=1}^n a_i e_i \otimes e_i $$
with $a_i \geq 0$.