The question goes as follows:
Let there be $\xi \in T_2(V)$ Where $V = M_n(\mathbb C))$:
$\xi (A,B) = n*tr(AB)-tr(A)tr(B).$
Find $dim(V_{\bot\xi})$.
$( V_{\bot\xi} = \{ A \in M_n(\mathbb C) | \xi(A,B) = 0, \forall B\in M_n(\mathbb C) \})$
I was struggling to find the dimension, this is what I tried to do:
First we mark $A=(a_{ij}), B = (b_{ij})$. Then, we want to find all $A \in M_n\mathbb C$ that $n*tr(Ab)=tr(A)tr(B)$ for every $B\in M_n\mathbb C$.
Since $tr(AB) = \sum_{i=1}^n \sum_{j=1}^n a_{ij}b_{ji}$, we get that:
$$n * \sum_{i=1}^n \sum_{j=1}^n a_{ij}b_{ji} = \sum_{i=1}^na_{ii}*\sum_{j=1}^nb_{jj}$$
$$\sum_{i=1}^n \sum_{j=1}^n na_{ij}b_{ji} = \sum_{i=1}^n\sum_{j=1}^na_{ii}b_{jj}$$ but I don't know how to continue from here. Am I on the right track? Is there any other intuitive way?
Thanks!
Hint Let $(E^a{}_b)$ denote the standard basis of $M_n(\Bbb C)$. The condition implies the $n^2$ conditions $$\xi(A, E^a{}_b) = 0 .$$ Write these conditions in terms of the entries of $A$.