Let W be a subspace of V. If dim W = dim V, then V = W Proof:

Answer 1

To prove that V = W, we need to show that every vector in V is also a vector in W, and vice versa.

Let's assume that dim V = dim W = n.

Since dim W = n, we can find a basis {w1, w2, ..., wn} for W which consists of n linearly independent vectors.

Since dim V = n, we can find a basis {v1, v2, ..., vn} for V which consists of n linearly independent vectors.

Now, let's take an arbitrary vector v in V. Since {v1, v2, ..., vn} is a basis for V, we can express v as a linear combination of these vectors. That is, v = a1v1 + a2v2 + ... + anvn, where a1, a2, ..., an are scalars.

Since V is a subspace of V, every linear combination of vectors in V is also in V. Therefore, v = a1v1 + a2v2 + ... + anvn is also a vector in V.

Now, we need to show that v is also a vector in W. Since {w1, w2, ..., wn} is a basis for W, we can express v as a linear combination of these vectors as well. That is, v = b1w1 + b2w2 + ... + bnwn, where b1, b2, ..., bn are scalars.

Since {v1, v2, ..., vn} is a basis for V, any vector in V can be expressed as a linear combination of these vectors. In particular, v can be expressed as a linear combination of v1, v2, ..., vn. Therefore, we can rewrite the expression for v in terms of v1, v2, ..., vn: v = c1v1 + c2v2 + ... + cnvn, where c1, c2, ..., cn are scalars.

Since every vector in V can be expressed as a linear combination of v1, v2, ..., vn, and every vector in V can also be expressed as a linear combination of w1, w2, ..., wn, we can equate the two expressions for v: c1v1 + c2v2 + ... + cnvn = b1w1 + b2w2 + ... + bnwn.

Since the basis {v1, v2, ..., vn} is linearly independent, the only way for the two sides of the equation to be equal is for the coefficients to be equal. That is, ci = bi for i = 1, 2, ..., n.

Therefore, we have v = c1v1 + c2v2 + ... + cnvn = b1w1 + b2w2 + ... + bnwn = v.

Since v is an arbitrary vector in V, and we have shown that v is both a vector in V and in W, we can conclude that V = W.