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Can someone explain why this one (the top answer) is false: For vector spaces, dim(U+V)=dimU+dimV−dim(UV), so dim(U+V+W)=dimU+dimV+dimW−dim(UV)−dim(UW)−dim(VW)
by bobbyi 16y ago
Can someone explain why this one (the top answer) is false:
For vector spaces, dim(U+V)=dimU+dimV−dim(UV), so
dim(U+V+W)=dimU+dimV+dimW−dim(UV)−dim(UW)−dim(VW)+dim(UVW)
- xi 16y agoIt is explained in the comments: take three different lines in the plane, and you'll get 2=3.
- jules 16y agoPerhaps more enlightening is where the proof fails: dim(U+V) = dimU+dimV−dim(UV) dim(U+V+W) = dim(U+V)+dim(W) - dim((U+V)W) = dim(U)+dim(V)+dim(W) - dim(UV) - dim(UW+VW) <-- bzzt, wrong = dim(U)+dim(V)+dim(W) - dim(UV)-dim(UW)-dim(VW) + dim(UVW) The reason this step fails is that (U+V)W != UW+VW for example UW and VW can both be empty, but (U+V)W is not empty: U = {multiples of (1,0)} V = {multiples of (0,1)} Now U+V is the entire space R^2. If we take W={multiples of (1,1)} then VW={0} and UW={0}, but because U+V is the entire space we have (U+V)W = W = {multiples of (1,1)}.
- omaranto 16y agoI can't tell you how many times I've had to explain this to confused linear algebra students... Oh, and A minor correction: you mean "zero" instead of "empty".
- jules 16y agoYep, that's right. HN doesn't allow me to edit :(
- xi 16y agoWe could also notice that UW+VW is a subspace of (U+V)W, and therefore the following statement is valid: dim(U+V+W)<= dim(U)+dim(V)+dim(W) - dim(UV)-dim(UW)-dim(VW) + dim(UVW). So the result is weakened, but not entirely lost.