Let W be a subspace of V. If dim W = dim V, then V = W Proof:
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.
To prove that if dim W = dim V, then V = W, we need to show that every vector in V is also in W, and vice versa.
Let's assume that dim W = dim V. This means that the dimensions of the two subspaces, W and V, are equal.
Now, let's consider the basis of W, which is a set of linearly independent vectors that span W. Since dim W = dim V, the same number of linearly independent vectors are required to span V.
Let's call the vectors that form the basis of W as w₁, w₂, ..., wₙ, where n is the dimension of W and V.
Since the dimension of V is also n, we can choose n vectors from the basis of V, say v₁, v₂, ..., vₙ, which are linearly independent and span V.
Since both sets of vectors, {w₁, w₂, ..., wₙ} and {v₁, v₂, ..., vₙ}, span V, we can express any vector in V as a linear combination of them.
Now, let's consider an arbitrary vector in V, say v. We can express v as a linear combination of the vectors from the basis of V:
v = c₁v₁ + c₂v₂ + ... + cₙvₙ (Equation 1)
Since the vectors {w₁, w₂, ..., wₙ} also span V, v can also be expressed as a linear combination of them:
v = d₁w₁ + d₂w₂ + ... + dₙwₙ (Equation 2)
Now, to prove that V = W, we need to show that every vector in V is also in W, and vice versa.
First, we will show that every vector in V is in W. From Equation 2, we have expressed v as a linear combination of the vectors {w₁, w₂, ..., wₙ}. This means that v is in the span of W, which implies that v is in W.
Next, we will show that every vector in W is in V. From Equation 1, we have expressed v as a linear combination of the vectors {v₁, v₂, ..., vₙ}. This means that v is in the span of V, which implies that v is in V.
Since every vector in V is in W, and every vector in W is in V, we can conclude that V = W.
Therefore, if dim W = dim V, then V = W.
To prove the statement, "If dim W = dim V, then V = W," we need to show that if the dimension of subspace W is equal to the dimension of the vector space V, then V must be equal to W.
To begin the proof, let's consider an arbitrary vector v in V. Since V is a vector space, v must be expressible as a linear combination of vectors in V. Mathematically, we can write:
v = c1v1 + c2v2 + ... + cnvn,
where v1, v2, ..., vn are vectors that form a basis for V, and c1, c2, ..., cn are scalars.
Next, let's consider the vectors v1, v2, ..., vn. Since V is a vector space, these vectors are also expressible as linear combinations of the basis vectors of W, which we'll denote as w1, w2, ..., wn:
v1 = a11w1 + a12w2 + ... + a1nwn,
v2 = a21w1 + a22w2 + ... + a2nwn,
...
vn = an1w1 + an2w2 + ... + annwn,
where aij are scalars.
Substituting these expressions back into the equation for v, we have:
v = c1(a11w1 + a12w2 + ... + a1nwn) + c2(a21w1 + a22w2 + ... + a2nwn) + ... + cn(an1w1 + an2w2 + ... + annwn)
= (c1a11 + c2a21 + ... + cnan1)w1 + (c1a12 + c2a22 + ... + cnan2)w2 + ... + (c1a1n + c2a2n + ... + cnann)wn.
Now, notice that we have expressed the vector v as a linear combination of vectors in W. This implies that v is in W. Since v was an arbitrary vector in V, we can conclude that every vector in V is also in W, thus proving that V is a subset of W.
To show the reverse inclusion, we can make a similar argument, starting with an arbitrary vector w in W and showing that w is also in V. By the same logic as above, we can conclude that W is a subset of V.
Since both V is a subset of W and W is a subset of V, we can conclude that V = W, as desired. This completes the proof.