Let $S:V\to V$ and $T:V\to V$ be two linear transformations such that:
$T^{2} = S^{2} = 0, T\circ S + S \circ T = Id$.
Prove that $V= Ker(S)\oplus Ker(T)$.
What can I use to prove this? Honestly, I don't know where to start.
2026-03-27 07:20:08.1774596008
Given two linear transformations, prove that V is equal to the direct sum of the kernels.
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If you apply $T$ (and $S$, respectively) to your equality you get $$ TST=T,\ \ STS=S. $$ If you now pre-multiply by $S$ and $T$ respectively, you get $$ STST=ST,\ \ TSTS=TS. $$ So $P=ST$, $Q=TS$ are projections ( $P^2=P$, $Q^2=Q$), and $$ P+Q=ST+TS=I. $$ Now any $v\in V$ can be written as $$ v=Pv+(I-P)v, $$ where $Pv\in\ker S$, $(I-P)v=Qv\in\ker T$.