I want to show that for every projection $A^2=A$ we have that there exists a subspace $U_1 \subset ker(A)$ and $U_2$ such that $A|_{U_2} = id$ such that $V = U_1 \oplus U_2$. Does anybody here have a hint how to show this?
2026-04-03 11:48:17.1775216897
Projection and direct sum
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Hint: (1) For $x \in V$ we have $x = (x-Ax) + Ax$.
(2) What can you say about $A(x-Ax)$ and $A(Ax)$?