In the case of vectors in $\mathbb R^n$, it is quite simple to see that for any vector $x$, $$ v^Txx^Tv = (v^Tx)^2 \geq 0$$ so clearly the form $xx^T$ must form a positive semidefinite matrix.
But what happens in other spaces? I guess $x^T$ generalize to some dual vector. but what about $xx^T$? That must be some sort of outer product? I have experience from Hilbert spaces in physics mostly, and there one forms such outer product often using dual vectors. But it is not clear for me that it should work 'as usual' in rigged Hilbert spaces...
In what spaces would this outer product make sense?
Edit: I guess the general idea could be to
- take $x \in V$, a vector in a vector space $V$ over the field $\mathbb K$.
- produce a dual vector $x^* \in V^*$ via Riez representation
- take the tensor product between the two, denote it $T \triangleq x \otimes x^* \in V \otimes V^*$. It can be interpreted as a bilinear map $T:(q,w) \mapsto T(q,w)=(qx)(x^*w)$, from the tensor product of the dual and the primal vector space $t:V^*\otimes V \to \mathbb K$.
- Define a quadratic form $X$ through the action $X(v) = T(v^*,v)$. From that definition it follows that $X(v) = (v^*x)(x^*v)$. Again, through riez representation, that is a square $X(v) = (x^*v)^2 \geq 0$, so the quadratic form indeed must be positive semidefinite.
In this construction, I guess I only need that $V$ is a Hilbert space, so if that is the case, then 'All hilbert spaces' is an answer. Is this reasoning valid? Is there a formal theory on this type of construction?