Using Rank-Nullity Theorem, we give an explicit basis for the kernel of the trace map.
Let \(n\geq 2\) be a natural number. Consider \(n\times n\) matrices over a field \(F\text{,}\)\(M_n(F)\text{.}\) Let \(E_{ij}\in M_n(F)\) be the matrix with \((i,j)\)-th entry \(1\) and all other entries zero. We have observed that \(\{E_{ij}:1\leq i,j\leq n\}\) is a basis for \(M_n(F)\text{.}\) Recall that we defined the trace of map in Example 4.2.5 as follows.
For any \(\alpha\in F\) the trace of the matrix \({\rm diag}(\alpha,0,\ldots,0)\) is \(\alpha\text{.}\) Thus \(tr\) is surjective. Therefore, \(\Im(tr)=F\) and
Hence, \(\alpha_{ij}=0,\beta_i=0\text{.}\) We thus get \(\{E_{ij}:i\neq j\}\cup\{E_{11}-E_{ii}:2\leq i\leq n\}\) is linearly independent.
The cardinality of the set \(\{E_{ij}:i\neq j\}\cup\{E_{11}-E_{ii}:2\leq i\leq n\}\) is \((n^2-n)+(n-1)=n^2-1\text{.}\) Since the \(\dim_F\ker(tr)=n^2-1\) the set \(\{E_{ij}:i\neq j\}\cup\{E_{11}-E_{ii}:2\leq i\leq n\}\) is a maximal linearly independent subset of \(\ker(tr)\text{,}\) i.e., it is a basis of \(\ker(tr)\) (see Lemma 3.4.1 and Theorem 3.4.3).