Kernel of a Linear Map on A Tensor Product Suppose I have the linear maps $ l,k: V \otimes V \rightarrow V \otimes V$ defined by
$ l( e_{i_1} \otimes e_{i_2} ) = e_{i_1} \otimes e_{i_2} + e_{i_2} \otimes e_{i_1}$
and 
$ k( e_{i_1} \otimes e_{i_2} ) = e_{i_1} \otimes e_{i_2} - e_{i_2} \otimes e_{i_1}$
where $ (e_{i})_{i=1}^{n} $ is a basis of V.
What are  the kernels of these two maps? 
Or at least an element in both their kernels... other than zero.

Here's one way to think about it: If $V$ (over, say, $\mathbb{F}$) is finite-dimensional, say, $\dim V = n$, then given a basis $(e_a)$ of $V$, we may identify $V \otimes V$ with the set of $n \times n$ matrices, so that the simple tensor $e_a \otimes e_b$ is identified with the matrix $E_{ab}$ whose $(a, b)$ entry is $1$ and for which all other entries are $0$, and so that $(E_{ab})$ is a basis for $V \otimes V$. Under this identification, the maps $l$ and $k$ act on basis elements by
$$l: E_{ab} \mapsto E_{ab} + E_{ba}$$
and
$$k: E_{ab} \mapsto E_{ab} - E_{ba}.$$
Now, we can write the general element $S \in V \otimes V$ as a matrix $\sum_{a,b} S_{ab} E_{ab}$, whose $(a, b)$ entry is $S_{ab}$. So,
$$l(S)
= l\left(\sum_{a, b} S_{ab} E_{ab}\right)
= \sum_{a, b} S_{ab} l(E_{ab})
= \sum_{a, b} S_{ab} (E_{ab} + E_{ba})
= \sum_{a, b} S_{ab} E_{ab} + \sum_{a, b} S_{ab} E_{ba}.$$
The first term on the right is $S$, and the second can be written as $\sum_{a, b} S_{ba} E_{ab}$, which the matrix whose $(a, b)$ entry is $(S_{ba})$, namely, the transpose ${}^t S$ of $S$. So, in terms of the matrix identification $l$ is just the map
$$l(S) = S + {}^t S$$
and similarly
$$k(S) = S - {}^t S.$$
So, $S$ is in the kernel of $l$ iff $S = -{}^t S$, that is, iff $S$ is antisymmetric. Similarly, $S$ is in the kernel of $k$ iff $S = {}^t S$, that is, if $S$ is symmetric. In both cases, we can apply the same terminology to the $2$-tensors in $V \otimes V$.
Moreover, each of the images is exactly the kernel of the other map: For example, if $S \in \ker l$, i.e., if $S$ is antisymmetric, then $k(S) = S - {}^t S = S - (-S) = 2 S$, and hence $S = \frac{1}{2} k(S)$.
This shows that we can actually decompose any $2$-tensor uniquely as a sum of a symmetric tensor and an antisymmetric tensor, and that this decomposition is given by
$$S = \frac{1}{2} l(S) + \frac{1}{2} k(S),$$
that is, $\frac{1}{2} l$ and $\frac{1}{2} k$ are respectively the vector space projections from $V \otimes V$ onto the subspaces of symmetric and antisymmetric tensors.
