If U is a proper subspace of a finitedimensional vector spac

If U is a proper subspace of a finite-dimensional vector space V, show that the dimension of U is less than the dimension of V.

Let U be a proper subspace of a finite dimensional vector space V.

As every vector space has a basis , let {u[1],u[2],...u[k]} be a basis for U.

As U is a proper subspace of V, there exists a vector v in V but not in U.

Claim: {u[1],u[2],...u[k],v} are linearly independent.

Let a[1]u[a]+a[2]u[2]+........a[k]u[k]+ cv =0

c cant be zero as that would mean {u[1],u[2],...u[k]} are linearly dependent , contradiction to the fact that

they form a basis (for U).

So c is not zero and dividing by c we obtain v as a linear combination of the u[i]s ,which means v is in U,

contrary to the assumption.

So {u[1],u[2],...u[k],v} are linearly independent.

Thus dimension V is at least dim U +1.

Hence the result