non-degeneracy on tensor product Let $(\cdot, \cdot)$ be a non-degenerate inner product on a finite dimensional vector space $V$ over a field $F$ of characteristic $0$. Define the inner product on $V \otimes V$ by $$(v \otimes w, v' \otimes w') = (v,v') \cdot (w,w')$$ and extend linearly. Is it necessarily the case that the inner product is non-degenerate on $V \otimes V$?
 A: Yes, it is... at least if $V$ is finite-dimensional. (I don't have the time to think about the other case; but I feel that it should be easily reducible to the former if you have an appropriate definition of "nondegenerate".)
Pick an orthogonal basis $\left(e_1,e_2,\ldots,e_k\right)$ for the inner product on $V$. (Such a basis exists, since the inner product is nondegenerate.) Notice that $\left(e_i,e_i\right) \neq 0$ for all $i$ (again because the inner product is nondegenerate). Then, $\left(e_i\otimes e_j\right)_{\left(i,j\right)\in\left\{1,2,\ldots,n\right\}^2}$ is an orthogonal basis for the inner product on $V \otimes V$ (and, again, the inner product of each of its elements with itself is nonzero). Hence, the latter inner product is nondegenerate.
The "characteristic $0$" condition can be replaced by the weaker requirement "characteristic $\neq 2$". More generally, if $F$ is a field of characteristic $\neq 2$, and if $V$ and $W$ are two finite-dimensional $F$-vector spaces equipped with nondegenerate symmetric bilinear forms $f$ and $g$ (I am uncomfortable with the word "inner product", since different people use it in different meanings), then the symmetric bilinear form $h$ on $V \otimes W$ determined by the rule
$$
h\left(v \otimes w, v^\prime \otimes w^\prime\right) = f\left(v, v^\prime\right) g\left(w, w^\prime\right)
$$
is nondegenerate. Again, the proof proceeds as before: Pick orthogonal bases and tensor.
Actually, we can go into greater generality: We can lift both the symmetry and the "characteristic $\neq 2$" conditions. We have the following fact:

Let $F$ be any field. Let $V$, $V^\prime$, $W$ and $W^\prime$ be four finite-dimensional $F$-vector spaces. Let $f : V \times V^\prime \to F$ and $g : W \times W^\prime \to F$ be two nondegenerate bilinear forms. (A bilinear form $k : U \times U^\prime \to F$ is said to be nondegenerate if and only if the two $F$-linear maps $U \to \left(U^\prime\right)^\ast$ and $U^\prime \to U^\ast$ induced by it (by "currying") are isomorphisms.) Then, the bilinear form $h : \left(V \otimes W\right) \times \left(V^\prime \otimes W^\prime\right) \to F$ determined by
  $$
h\left(v \otimes w, v^\prime \otimes w^\prime\right) = f\left(v, v^\prime\right) g\left(w, w^\prime\right)
$$
  is nondegenerate.

The proof of this fact is particularly simple: The map $V^\prime\otimes W^\prime \to \left(V\otimes W\right)^\ast$ induced by $h$ is identical with the tensor product of the map $V^\prime \to V^\ast$ induced by $f$ with the map $W^\prime \to W^\ast$ induced by $g$ (provided that we canonically identify $ \left(V\otimes W\right)^\ast$ with $V^\ast \otimes W^\ast$); therefore, it is an isomorphism whenever $f$ and $g$ are isomorphisms. A similar argument deals with the other map.
