= Solution
The map $T\mapsto T^{\mathsf T}A+AT$ is linear, so its <kernel> $L$ is a <vector subspace> of the <matrix algebra>. If $T_1,T_2\in L$, use $T_i^{\mathsf T}A=-AT_i$ to compute
$$
\begin{aligned}
[T_1,T_2]^{\mathsf T}A
&=(T_2^{\mathsf T}T_1^{\mathsf T}-T_1^{\mathsf T}T_2^{\mathsf T})A\\
&=-T_2^{\mathsf T}AT_1+T_1^{\mathsf T}AT_2\\
&=AT_2T_1-AT_1T_2=-A[T_1,T_2].
\end{aligned}
$$
Thus $L$ is closed under the <commutator>. This bracket is bilinear and satisfies $[T,T]=0$. Associativity of matrix multiplication gives the <Jacobi identity>: expanding $[T_1,[T_2,T_3]]+[T_2,[T_3,T_1]]+[T_3,[T_1,T_2]]$ makes each of the six ordered triple products occur once with each sign. Therefore \b[$L$ is a Lie algebra]. No nonsingularity or symmetry assumption on $A$ was needed.
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