$\ker(β : V \otimes V → k)$ contains the subspace ${V}^{ \perp}\otimes V +V \otimes {V}^{\perp}$, I found this problem and I have been trying to solve it but I have some questions:
1. If a linear map is from a tensor product to a field, the map isn't suppose to be linear and not bilinear? I know that a bilinear map is from $V \times  W$  and it has to be linear in its two components.
2. What is symmetry properties? 
The problem 
Let $k$ denote the field of scalars (either Reals or Complex numbers). Suppose that
$β : V \otimes V \to k$ is a bilinear form, symmetric or anti-symmetric. 
Let
${β}_{1}: V \to {V}^{*}$ 
denote the map defined by ${β}_{1}(a)(b) = β(a, b) $ and let ${V}
^{\perp} :=
\ker({β}_{1}).$
Show that $\ker(β : V \otimes V \to k)$ contains the subspace ${V}^{
\perp}\otimes V +V \otimes {V}^{\perp}$,
hence the bilinear form $β$ induces a bilinear form $ {β}_{2} : V/{V}^{\perp} \otimes V/{V}^{ \perp} \to k$
with the same symmetry properties as those of $β$ and that ${β}_{2}$ is nondegenerate.
 A: 1) Yes, you are right, the original problem is a bit sloppy with notation. For every bilinear $\beta$ map $V \times W \to X$ there is an associated linear map $\beta': V \otimes W \to X$ which on `pure tensors' $v \otimes w$ is defined by $\beta'(v \otimes w) = \beta(v, w)$ and then extended to all of $V \otimes W$ by demanding that $\beta'$ is linear. We should then use bilinearity of the original map to verify that this is indeed well defined, i.e. that whenever some pure tensor $v \otimes w$ can be written as a linear combination of more than 1 other pure tensors both expressions for the same element of $V \otimes W$ produce the same element in $X$ when fed into $\beta'$.
The staters of the problem use $\beta$ and $\beta'$ completely interchangeably, hopping freely from one map to the other as if they were the same thing. This is somewhat justified by the fact that not only is there for every bilinear map $\beta$ with domain $V \times W$ an associated linear map $\beta'$ with domain $V \otimes W$, but also for every linear map $\gamma: V \otimes W \to X$ we can easily find a bilinear map $\beta$ so that $\beta' = \gamma$. (Check this for yourself.) So while sloppy, the staters of the problem are not entirely incorrect, it is loosely comparable to people using 'matrix' and 'linear map' as interchangeable.
2) Symmetric means that $\beta(v, w) = \beta(w, v)$ or equivalently $\beta(v \otimes w) = \beta(w \otimes v)$ and anti-symmetric means $\beta(v, w) = -\beta(w, v)$ or equivalently $\beta(v \otimes w) = -\beta(w \otimes v)$. The claim of the problem is that if $\beta$ is symmetric then so is $\beta_2$ and that if $\beta$ is anti-symmetric, then so is $\beta_2$.
