4 ms·
Perhaps 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)
by jules 16y ago
Perhaps 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.