Dimension of the basis $\{x \otimes y + y \otimes x\}$

Question

I'm trying to prove that the annihilator of $I = \left<x \otimes y - y \otimes x \right>$ is $\left<x \otimes y + y \otimes x \right>$. To do this I am trying to compare dimensions. So if $\{x_i \otimes x_j -x_j \otimes x_i \mid x_i,x_j \in V\}$ where $V$ is an n dimensional vector space is a basis for $I$, which has dimension $ \binom {n} {2}$, then $\{x^*_i \otimes x^*_j + x^*_j \otimes x^*_i \mid x^*_i,x^*_j \in V^*\}$ has dimension $ \binom {n+1}{2}$ and thus forms a basis for $I^{\perp}$. I'm struggling to work out the dimension of $\{x^*_i \otimes x^*_j + x^*_j \otimes x^*_i \mid x^*_i,x^*_j \in V^*\}$ and am not too sure how to count the elements.

Answer 1

How can you choose $x_i^*,x_j^*$ among the elements of the dual basis? Well, they can either be different, which gives you $\binom{n}{2}$ elements, or you can have $x_i^*=x_j^*$, giving you $n$ more possibilities, so $$\dim\{x^*_i \otimes x^*_j + x^*_j \otimes x^*_i \mid x^*_i,x^*_j \in V^*\} = \binom{n}{2} + n.$$ Now, $$\binom{n}{2} + \left(\binom{n}{2} + n\right) = n(n-1) + n = n^2 = \dim V^{\otimes2},$$ and you are done.