Under what conditions does $S(U) + T(U) = (S+T)(U)$ for linear operators S,T? Let $V$ be a finite dimensional (real) vectors space, $U \leq V$ a linear subspace, and $S, T: V \rightarrow V$ be two linear maps. Then $(T+S)(U) \subset T(U) + S(U)$ holds clearly. What are sufficient and necessary conditions under which we have an inclusion in the opposite direction, ie. when is $(T+S)(U) = T(U) + S(U)$?
Is there a way to formulate this in terms of the inner product defined on operators as $\langle S, T\rangle = Tr(A^T B)$?
 A: Consider the natural maps $\phi\colon V\to V\times V$ sending $v\mapsto (v, v)$ and the map $\psi\colon V\times V\to V$ sending $(v_1,v_2)\mapsto Tv_1+Sv_2$
Then $(T+S)(U)=(\psi\circ \phi)(U)$ while
$T(U)+S(U)=\psi(U\times U)$
We know that $ (T+S)(U)\subseteq T(U)+S(U)$ so that the two spaces are equal if and only if they have the same dimension:
$\dim ((T+S)(U))=\dim(U)-\dim(\ker (\psi\circ \phi))$
while
$\dim (T(U)+S(U))=2\dim(U)-\dim(\ker(\psi))$
(Here $\psi$ and $\phi$ are restricted on $U\times U$ and $U$ respectively)
Thus
$\dim(\ker(\psi))-\dim(\ker (\psi\circ \phi))=2 \dim(U)-\dim(U)=\dim(U)$
Remember that
$\ker(\psi)=\{(v_1,v_2)\in U\times U: Tv_1=-Sv_2\}$
and so it’s the fibered product of $T$ and $-S$, $U\times_{T,-S} U$
and
$\ker(\psi\circ \phi)=\{ v\in U : Tv=-Sv\}$
whose image with respect to $\phi$ is the diagonal $\Delta$ intersected with $ U\times_{T,-S} U$
Another equivalent condition is that
$\dim(\ker(\psi)/\phi(\ker(\psi\circ \phi)))=\dim(U)$
and
$\frac{U\times_{T,-S} U}{\left(\Delta\cap U\times_{T,-S} U\right)}=\ker(\psi)/\phi(\ker(\psi\circ \phi))\cong U$
