Prove that there exists $\beta \in L^1(V)$ such that $T^s=\alpha \otimes \beta + \beta \otimes\alpha$ Assume that $V$ is a vector space, $T\in L^2(V)$ is a $2$-tensor and $\alpha \in L^1(V)=V^*$ is a nonzero linear transformation.
For each $v \in Ker(\alpha)$ we have $T(v,v)=0$.  
Prove that there exists $\beta \in L^1(V)$ such that $T^s=\alpha \otimes \beta + \beta \otimes\alpha$.   
Note 1: $L^k(V)$ is the set of all $k$-tensors on $V$.  
Note 2: My problem is "how to find that $\beta$". I don't know where to start.  
Note 3: The defenition of $T^s$ is $T^s(u,v):=\frac{T(u,v)+T(v,u)}{2}$
 A: I'll assume that $\operatorname{char} \mathbb{F} \neq 2$. Note that $T|_{\ker \alpha}$ is alternating and so $T^s|_{\ker \alpha} = 0$. More explicitly, if $u, v \in \ker \alpha$ then
$$ 0 = T(u + v, u + v) = T(u,u) + T(u,v) + T(v,u) + T(v,v) = 2T^s(u,v) $$
so $T^s(u,v) = 0$. If $\alpha = 0$ then we see that $T^s = 0$ so we can take $\beta = 0$. If not, choose $w$ such that $\alpha(w) = 1$ and then $V = \ker \alpha \oplus \operatorname{span} \{ w \}$. Define a linear functional on $V$ by
$$ \beta(u) = \begin{cases} T^s(u,w) & u \in \ker \alpha, \\
\frac{T^s(w,w)}{2} & u = w. \end{cases} $$
Then
$$
(\alpha \otimes \beta + \beta \otimes \alpha)(u_1,u_2) = \\ \alpha(u_1) \beta(u_2) + \beta(u_1) \alpha(u_2) =  \begin{cases}
0 = T^s(u_1,u_2) & u_1, u_2 \in \ker \alpha, \\
\beta(u_1) = T^s(u_1,u_2) & u_1 \in \ker \alpha, u_2 = w, \\
\beta(u_2) = T^s(u_2,u_1) = T^s(u_1,u_2) & u_1 = w, u_2 \in \ker \alpha, \\
2\beta(w) = T^s(u_1,u_2) & u_1 = u_2 = w
\end{cases} $$
which shows that $\alpha \otimes \beta + \beta \otimes \alpha = T^s$.
