Basis for kernel of trace map Let $T: M_{n \times n}(F) \rightarrow F$  be defined by $T(A) = tr(A)$. 
I want to find out what a basis is for the kernel $N(T)$ of this linear map. I know $tr(A) = \sum_{i=1}^{n} A_{ii}$. 
I also know the dimension of the kernel has to be $n^2 - 1$, because $rank(T) = 1$ and $dim(M_{n \times n} (F)) = n^2$. But how do I find a suitable basis for the kernel? 
 A: Let $\delta_{ij}$ be the matrix with a $1$ in $(i,j)$ and $0$s elsewhere. Then if $i \neq j$, this matrix has zeroes on the entire diagonal; so $\text{Tr}(\delta_{ij}) = 0$. We now have $n^2-n$ linearly independent terms; we just need to find $n-1$ more.
So far we've ignored the diagonal, so let's find elements of the kernel of $\text{Tr}$ that only have elements on the diagonal. One set is $\delta_{11} - \delta_{ii}$ for $n \geq i > 1$; you should check that, indeed, $$\{\delta_{ij} : 1 \leq i \neq j \leq n \} \cup \{\delta_{11}-\delta_{ii} : 1 < i \leq n \}$$ is a linearly independent set. Then you're automatically done, as you know that $N(\text{Tr})$ has dimesnion at least $n^2-1$; but it can't have dimension $n^2$, as $\text{Tr}(\delta_{11}) = 1 \neq 0$. So this is a basis for $N(\text{Tr})$.
A: Let $E_{ij}$ denote the matrix whose entries are all zero except for the $i,j$ entry, which is a $1$.  Clearly, these matrices form a basis of $M_{n\times n}(F)$.
Consider the matrices $E_{ij}$ with $i \neq j$ and $(E_{ii} - E_{nn})$ with $i<n$.
