◦
E1, ..., EpRn
E1∩E2={0}E1+E2
E1∩E2∩E3={0}E1+E2+E3
E1+...+Epk∈ {2,3, ..., p}
(E1+... +Ek−1)∩Ek={0}
f1:R3→R2f1(x, y, z) = (x, y)
f2:R2→R4f2(x, y) = (−αx, y +√2x, 0, β(2x−y)), α, β ∈R
f3:R2→R2f3(x, y) = (x, −1)
f4:R3→Rf4(x, y, z) = ax +by +cz, a, b, c ∈R
f5:R2→R2f5(x, y) = (|x|, y)
f6:R2→R3f6(x, y) = 3(x, y, x −y)
f1, f4f6
X Y f :X→Y g :Y→X
f◦g=idY
f g
g◦f=idXf g
E F Rnf:E→F
Im f ={f(x), x ∈E}fIm f =F
Ker f ={x∈E|f(x)=0}fKer f =
{0}
g:R3→R3g(x, y, z) = (y, z, x)
g
g2=g◦g g3g4g11
g g−1