Let U and W be the subspaces of V and that dim U = 4,
dim W = 4 and dim V = 6. Then the possible dimensions of
U W
(a) 1,2
(b) 2.3
(c) 3.4
(d) 4,5
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Step-by-step explanation:
Notice that U∩W is a subspace of U and W so
dim(U∩W)≤4
but since dim(U∩V)=4⟺V=U∩V=U
hence
dim(U∩W)≤3
if dim(U∩W)=0 then the sum U+W is direct but in this case
dim(U+W)=dimU+dimW>6
which's impossible.
if dim(U∩W)=1. Let U∩W=span(e1) and let (e1,u2,u3,u4) a basis for U and (e1,w2,w3,w4) a basis for W hence we see easily that dim(U+W)=7>6 which's impossible.
The cases dim(U∩W)=2 or dim(U∩W)=3 are possible using the same method as on the last point.
Conclusion The possible values of dim(U∩W) are 2 and 3.
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