 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 $T_i^{\mathsf{T}}A=-A{T}_i$ to computeThus $L$ 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 $L$ is a Lie algebra. No nonsingularity or symmetry assumption on $A$ was needed.