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Section 4.6 A basis for the kernel of trace map

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.
\begin{equation*} tr\colon M_n(F)\to F\quad\text{given by}\quad(a_{ij})\mapsto\sum_i a_{ii}. \end{equation*}
By Rank-Nullity Theorem (Theorem 4.5.3),
\begin{equation*} \dim_F\ker(tr)=\dim_FM_n(F)-\dim_F\Im(tr). \end{equation*}
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
\begin{equation*} \dim_F\ker(tr)=n^2-1. \end{equation*}
Note that \(E_{ij}\in\ker(tr)\) for every \(i\neq j\) and \(E_{11}-E_{ii}\in\ker(tr)\) for every \(2\leq i\leq n\text{.}\)
Suppose that there exists \(\alpha_{ij},\beta_i\in F\) such that
\begin{equation*} \sum_{i\neq j}\alpha_{ij}E_{ij}+\sum_{i\neq 1}\beta_i(E_{11}-E_{ii})=0. \end{equation*}
In other words the above equation is
\begin{equation*} \begin{pmatrix}\sum_{i=2}^{n}\beta_i\amp\alpha_{12}\amp\alpha_{13}\amp\cdots\amp\alpha_{1n}\\\alpha_{21}\amp-\beta_2\amp\alpha_{23}\amp\cdots\amp\alpha_{2n}\\\vdots\amp\vdots\amp\vdots\amp\ddots\amp\vdots\\\alpha_{n1}\amp\alpha_{n2}\amp\alpha_{n3}\amp\cdots\amp -\beta_n\end{pmatrix}=0. \end{equation*}
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).