Suppose that V is a vector space and dimV 3 and W is a subs

Suppose that V is a vector space and dim(V) = 3 and W is a subspace of V. Prove directly (without simply quoting a theorem) that W must have finite dimension.

Solution

Suppose V is a vector space and dim(V)=3

Let the set of vectors {₁, ₂, ₃} be a basis of V. Then any vector in V can be expressed as linear combination of ₁, ₂, ₃

Suppose that W is a subspace of V and let is a vector in W

Then is also a vector in V and thus can be expressed as linear combination of ₁, ₂, ₃

Now If the set of vectors ₁, ₂, ₃ are independent in W then this set {₁, ₂, ₃} is a basis for W and hence dim(W) =3 which is finite.

If the set of vectors ₁, ₂, ₃ are dependent in W then a subset of {₁, ₂, ₃} forms a basis of W and in this case dim(W) < 2, which is also finite.

Therefore in any case W is finite dimensional