# Characterising the cone of elements whose inner product with $v \otimes v$ is non-negative

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.

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 $$a_i \geq 0$$.

• In other words, your condition puts a restriction only on the symmetric part of $\sigma$ leaving the alternating part completely arbitrary. – levap Aug 18 at 14:17
• This is similar to how the condition $\left< Av, v \right> \geq 0$ doesn't imply that $A$ is symmetric as this condition doesn't depend on the antisymmetric part of $A$ which can be arbitrary while the symmetric part can be written as $\sum_{i=1}^n a_i e^i \otimes e_i$ with $a_i \geq 0$. – levap Aug 18 at 14:35
• I want to add a specific counterexample: take $V = \mathbb{R}^2$ and $\sigma = e_1 \otimes e_1+e_1 \otimes e_2 + e_2 \otimes e_2$. Take any $v \in V$, we have $\langle \sigma, v \otimes v \rangle = (v_1)^2 + v_1v_2 +(v_2)^2$ which you can write as either $(v_1-v_2)^2 + 3v_1v_2$ or $(v_1+v_2)^2 -v_1v_2$ to see that the inner product is positive for all $v_1,v_2 \in \mathbb{R}$. Secondly you cannot represent $\sigma$ in the way you want since every linear combination of tensors of the form $v \otimes v$ is symmetric and $\sigma$ is not. – abcdef Aug 18 at 14:58
• I would just go with the counterexample $\sigma = e_1 \otimes e_2 - e_2 \otimes e_1$ as $\left< \sigma, (v_1 e_1 + v_2 e_2) \otimes (v_1 e_1 + v_2 e_2) \right> = v_1 v_2 - v_2 v_1 \equiv 0.$ and, as you write, it is not a linear combination of tensors of the form $v \otimes v$ because it is alternating and not symmetric. – levap Aug 18 at 15:10
• @levap Thanks! Just to be sure: The fact that "we can find an orthonormal basis such that..." is essentially the theorem that every real symmetric matrix is orthogonally diagonalizable, right? – Asaf Shachar Aug 18 at 15:15