Suppose U1 Um are finitedimensional subspaces of V Prove tha

Suppose U_1,..., U_m are finite-dimensional subspaces of V. Prove that U_1 +... + U_m is finite-dimensional and dim (U_1 +... U_m) lessthanorequalto dim U_1 +... + dim U_m

Solution

Given that U1, U2...Um are finite dimensional subspace of V

If U1, U2 ...Um are mutually exclusive and exhaustive then V would have been partitioned into exhaustic m disjoint subspaces.

In that case dim (U1+U2+...Um) = dim V = dim U1+dim U2+...dim Um

In other cases where pairwise intersection of Ui\'s is not a null set

for example take two sets U1 and U2 where Intersection consists of atleast one element.

The common element can be represented as a linear combination of both bases of U1 and U2.

So dim(U1+U2) < dim U1+dim U2

This can be extended to m subspaces by induction

Let dim (u1+U2+...+Um) <DimU1+dimU2+...dimUm where intersection is not a null set

Consider another space Um+1 which is having intersection with atleast one Ui

Then dim (U1+....Um) + dim Um+1 > dim (U1+...+Um+1)

Hence proved by induction