Is there a nice argument that shows why $GL_{2}(K)$ acts on $\{X\in Mat_{2}(K)|trace(X)=0\}$ as $Sym^{2}\otimes\det^{-1}$ if $char(K)\neq 2$? Is there a nice argument (other than brute-force calculation) that shows that $\operatorname{GL}_{2}(K)$ acts on $\{X\in\operatorname{Mat}_{2}(K)|\operatorname{trace}(X)=0\}$ as $\operatorname{Sym}^{2}\otimes\det^{-1}$ if $\operatorname{char}(K)\neq 2$?
By brute force I mean I explicitly wrote down how the generators of GL_{2}(K) act on a basis of $\operatorname{Sym}^{2}$ and built a linear system of equations that mimicked that action on a suitable basis of $\{X\in\operatorname{Mat}_{2}(K)|\operatorname{trace}(X)=0\}$, yielding
\begin{equation*}
\begin{matrix}\left(\begin{matrix}1\\0\end{matrix}\right)^{\otimes 2}\longmapsto\left(\begin{matrix}0 & 2w\\0 & 0\end{matrix}\right)\\
\left(\begin{matrix}1\\0\end{matrix}\right)\otimes\left(\begin{matrix}0\\1\end{matrix}\right)\longmapsto
\left(\begin{matrix}w & 0\\0 & -w\end{matrix}\right)\\
\left(\begin{matrix}0\\1\end{matrix}\right)^{\otimes 2}\longmapsto
\left(\begin{matrix}0 & 0\\-2w & 0\end{matrix}\right)\end{matrix}\end{equation*}
for $w\in K^{\times}$, as the only possible isomorphism.
Also, can this be generalized to dimensions higher than $2$?
(If $\operatorname{char}(K)=2$, the statement remains true if and only if $K=\mathbb{F}_{2}$, but with a different isomorphism than the one given.)

*

*$GL_2(K)$ acts on $X$ through $$g\cdot M = gM g^{-1}$$


*$Sym^2 = K^2\otimes_K K^2/Kv$ where $v=(1,0)\otimes (0,1)-(0,1)\otimes (1,0)$.
The $K$-vector space isomorphism $K^2\otimes_K K^2\to M_2(K)$ gives the $GL_2(K)$-module isomorphism $$Sym^2 \cong M_2(K)/K A$$ where $A = \pmatrix{0&1\\-1&0}$ and $GL_2(K)$ acts on $B\in M_2(K)$ in the RHS through $$g\cdot B = g B g^\top$$ Note that $g\cdot A = \det(g) A$


*We have $\det(g) g^{-1} = (-AgA)^\top$ so
$g^\top = -A\det(g) g^{-1}A$
Whence $$g\cdot BA^{-1} = g B \det(g) g^{-1}A^{-1}$$
From which the isomorphism $Sym^2\otimes \det^{-1} \cong X$ is clear.
