To prove that a subspace of a finite-dimensional vector space is finite-dimensional, you can use the following argument:
Let V be a finite-dimensional vector space, and let W be a subspace of V. To show that W is finite-dimensional, we need to demonstrate that there exists a finite basis for W.
First, note that since W is a subspace of V, it inherits the vector space properties from V. This means that W is closed under addition and scalar multiplication.
Now, consider a basis for V, let's call it B = {v1, v2, ..., vn}, where n is the dimension of V. Since B is a basis for V, it spans V and is linearly independent.
Next, we need to show that there exists a finite subset of B that forms a basis for W. To do this, consider the set S = B ∩ W, which consists of the elements of B that are also in W.
Now, we claim that S is a basis for W. To prove this, we need to show that S spans W and is linearly independent.
Spanning property: Let w be any vector in W. Since W is a subspace of V, w is also in V. Since B is a basis for V, w can be expressed as a linear combination of the vectors in B: w = c1v1 + c2v2 + ... + cnvn, where c1, c2, ..., cn are scalars. Now, since w is in W, it must also be in the intersection of B and W, which means w is in S. Thus, we have shown that S spans W.
Linear independence: Suppose there exists a linear combination of vectors in S that equals the zero vector: a1s1 + a2s2 + ... + aksk = 0, where a1, a2, ..., ak are scalars, and s1, s2, ..., sk are vectors in S. Since each si is in S, it is also in W. Therefore, the linear combination above is a linear combination of vectors in W that equals zero. However, since B is linearly independent, the only way for this linear combination to equal zero is if all the coefficients a1, a2, ..., ak are zero. Thus, we have shown that S is linearly independent.
Since S is a finite set that both spans W and is linearly independent, it forms a basis for W. Therefore, W is finite-dimensional.
In conclusion, every subspace of a finite-dimensional vector space is itself finite-dimensional.