Why a projection is singular and an identity mapping?

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I got this question on one of my linear algebra assignments. I don't understand how to prove any of the points asked, theoretically it seems pretty trivial it doesn't even need a proof.

Let E be a linear operator on R^n for which $E^2 = E$. (Such an operator is called a projection.) Let U be the image of E, and let W be the kernel of E. Prove:

  • If $u ∈ U$, then $E(u) = u$, i.e., E is the identity mapping on U.
  • If $E \not= I$, then E is singular, i.e., $E(v) = 0$ for some $v \not= 0$.
  • $R^n = U ⊕ W$