this is not always TRUE.
if U = W, then dim(U) = dim(W) = n -1, but dim(U∩W) = n-1.
you can't prove things that are not true.
then it's a different story.
since W and U are distinct subspaces, say with bases B and B', respectively, then B' must contain a vector not in span(B). for if not, then U ⊆ W, in which case dim(U) = dim(W) forces U = W.
hence dim(U+W) = dim(span(B U B')) = n (glossed over a bit: if we call this vector u_{j}, with B = {w_{1},...,w_{n-1}}, B' = {u_{1},...,u_{n-1}}, we need to show:
C = BU{u_{j}} is linearly independent. however, if:
c_{1}w_{1}+...+c_{n-1}w_{n-1}+c_{n}u_{j} = 0, we can distinguish 2 cases:
a) c_{n} = 0. in this case, the linear independence of the w_{i} forces c_{1} = ... = c_{n-1} = 0,
b) c_{n} ≠ 0. in this case, u_{j} = (-c_{1}/c_{n})w_{1}+...+(-c_{n-1}/c_{n})w_{n-1}, contradicting our choice of u_{j}.
we also have n-1 = dim(span(B)) < dim(span(B U {u_{j}}) ≤ dim(B U B') ≤ n).
therefore: dim(U∩W) = dim(U) + dim(W) - dim(U+W) = n-1 + n-1 - n = 2n - 2 - n = n - 2.
but for the converse one? So far, I only have a result that if K is a set consisting of all nonunits of R, then for any u, v in K, u+v won't be 1R. Does it help? Any suggestion?
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