trace function ($2\times2$) with ordered bases as linear transformation

Question

We got trace function as following:

$$\operatorname{tr}\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}=a+d$$

So now have to write down $[\operatorname{tr}]_{S_1,S_2}$, where ordered bases are the standard bases:

$$S_2=\left\{\begin{bmatrix} 1 & 0\\ 0 & 0\\ \end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix}, \begin{bmatrix} 0 & 0\\ 0 & 1\\ \end{bmatrix}\right\}, S_1=\{1\} $$

I'm pretty sure its $S_1=\{1\} $, not the $I$ (identity) matrix. Since we are doing transformation from Matrix to number ($4\dim$ to $1\dim$)

So the question is: "can the trace function be represented as a linear transformation?"

Please advice.

Answer 1

Let the basis elements be named $e_1, e_2, e_3, e_4$ in the order given in the OP. Then the matrix $\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ can be written as $ae_1 + be_2 + ce_3 + de_4$ = $\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$.

$ \text{tr}\begin{pmatrix}a & b\\ c & d\end{pmatrix} = a + d \Rightarrow \\ \text{tr}(ae_1 + be_2 + ce_3 + de_4) = 1a + 0b + 0c + 1d $

From this, the matrix representing $\text{tr}$ is $\boxed{(1\ 0\ 0\ 1)}$.

Answer 2

$tr$ is represented by the $1\times 4$ matrix $(1,0,0,1)$.