Let $V,W$ be two real vector spaces, and let $L_i:V\rightarrow W$, $i=1,\ldots,n$ be $n$ linear maps with distinct kernels $K_i$ of dimension $1$.
Consider the tensor product of these maps $L_1\otimes \ldots \otimes L_n: V\otimes \ldots \otimes V \rightarrow W \otimes \ldots \otimes W$. Let K denote the kernel of this map, and let $S^n(V)$ be the space of symmetric tensors of order n.
How can I prove that $K \cap S^n(V) = Span\{K_i \otimes \ldots \otimes K_i , i=1,...n\}$?
Thanks!
I can prove the equality under a stonger assumption:
The kernels $K_i$ are in direct sum inside $V$.
With this assumption, there is a basis $\{v_1, \ldots, v_n, v_{n+1}, \ldots, v_d\}$ of $V$ such that for $i=1,\ldots, n$, we have $v_i\in K_i$. The kernel $K$ is equal to $K_1\otimes V\otimes \cdots \otimes V + V\otimes K_2\otimes \cdots \otimes V + \ldots + V\otimes \cdots \otimes V\otimes K_n$ (this can be seen by using induction and this answer). This means that a basis of $K$ is $$\{ v_{i_1} \otimes \cdots \otimes v_{i_n} \ | \ i_j = j \text{ for some } j \}.$$
Now, let $x$ be an element of $K\cap S^n(V)$. By the above, $x$ has the form $$ x=\sum\lambda_{i_1, \ldots, i_n}v_{i_1} \otimes \cdots \otimes v_{i_n}, $$ where $\lambda_{i_1, \ldots, i_n}$ is non-zero only if $i_j=j$ for some $j$. The claim is that $\lambda_{i_1, \ldots, i_n}$ is non-zero only if $i_1=\ldots = i_n$.
Since $x\in S^n(V)$, for any permutation $\sigma$, we have that $\lambda_{i_1, \ldots, i_n} = \lambda_{i_{\sigma(1)}, \ldots, i_{\sigma(n)}}$. Assume that $\lambda_{i_1, \ldots, i_n}$ is non-zero, and let $J=\{j \ | \ i_j = j \}$. By the above, $J$ is not empty. If $J$ contained $2$ elements or more, then any permutation $\sigma$ fixing the elements not in $J$ and fixing no element in $J$ would be such that $\sigma(i_k)\neq k$ for all $k$, so $\lambda_{i_{\sigma(1)}, \ldots, i_{\sigma(n)}}=0$, a contradiction. Thus $J$ contains exactly one element, say $j$. Then applying any transposition of the form $\sigma=(k,j)$, we get that $\lambda_{i_{\sigma(1)}, \ldots, i_{\sigma(n)}}$ is non-zero. This implies that $i_{k} = i_{\sigma(j)} = j$. Since this is true for all $k$, we get that $\lambda_{i_1, \ldots, i_n} = \lambda_{j,\ldots, j}$. This proves the claim.
