The Dimension Theorem of a Vector Space Basis
Definition
In a vector space V over the field K, if there exists a basis B of V with a finite number of elements (dimension), then every other basis B' of V will have the same number of elements.
Proof
Consider a vector space V with a basis B consisting of n elements.
$$ B={v_1,...,v_n} $$
Assuming the contrary, there exists another basis B' with m elements, where m>n, i.e., with more elements than the previous basis B.
$$ B' = \{ v_1 ,..., v_n , v_{n+1} ,..., v_m \} $$ $$ with \:\: m>n $$
According to the basis completion theorem, I can add the additional m-n vectors of B' to B.
$$ {v_{n+1},...,v_m} $$
However, this leads to a contradiction.
If B is a basis of the vector space with n elements (dimension), then it is already a maximal set of linearly independent vectors.
By the theorem of linear dependency of vectors relative to the basis, any vector added to B must necessarily be linearly dependent on the previous ones.
Therefore, if B is a basis, then B' cannot be a basis of the vector space V because it contains m-n linearly dependent vectors.
This violates one of the fundamental requirements of vector bases.
In conclusion, the only possible solution to avoid this contradiction is as follows:
$$ m=n $$
Two bases of V must have the same number of elements, i.e., the same dimension, because if there exists a basis with n elements, then there cannot also exist a basis with m elements (where m>n).
Note. The opposite case can also be proven. If there is a basis with n elements, then there cannot exist a basis with fewer than n elements, as it would be incomplete. Therefore, it must be completed by adding the missing n-k linearly independent vectors.
Alternative Proof
Consider two bases of the vector space V
Initially assuming
- The first basis is a set of generators consisting of n vectors $$ \{ v_1, v_2, ... , v_n \} $$
- The second basis is a set of linearly independent vectors $$ \{ w_1, w_2, ... ,w_p \} $$
According to the theorem on linear independence, a set of linearly independent vectors of the vector space V always has a cardinality less than or equal to any set of generators of V.
Therefore, the basis with linearly independent vectors {w1,w2,...,wp} consists of a number of vectors less than or equal to any set of generators of V, including {v1,v2,...,vn}.
$$ p \le n $$
Moreover, initially assuming {v1,v2,...,vn} is a basis as well as a set of generators.
Thus, {v1,v2,...,vn} also has a number of vectors less than or equal to any other set of generators of V.
Since {w1,w2,...,wp} is a basis of V, it is also a set of generators of V.
Therefore, {v1,v2,...,vn} has a number of vectors less than or equal to {w1,w2,...,wp}.
$$ n \le p $$
Combining these two conditions leads to the conclusion that n equals p.
$$ \begin{cases} p \le n \\ \\ n \le p \end{cases} \Longleftrightarrow n = p $$
The two bases {v1,v2,...,vn} and {w1,w2,...,wp} have the same number of vectors n=p, meaning they have the same dimension.
Corollaries
From the dimension theorem, other corollaries can be derived.
Corollary 1
If the vector space V has dimension n, to obtain a basis B of the vector space, it is sufficient to find n linearly independent vectors v∈V.
Corollary 2
If the vector space V has a finite dimension n over the field K, then a vector subspace W of V has dimension n if and only if W=V.
$$ dim(W) = dim(V) \:\: if \: W = V $$ $$ dim(W) < dim(V) \:\: if \: W \ne V $$
Proof
Given a subspace W of the vector space V, we have
$$ W ⊆ V $$
$$ dim_k(W) = m $$
$$ dim_k(V) = n $$
The cardinality of the subspace W must necessarily be m≤n; otherwise, not all vectors would be linearly independent according to the theorem of vector dependency relative to the basis.
If dimk(W) = dimk(V), then the subspace W has a basis composed of m=n linearly independent vectors.
$$ B_w = \{ v_1 , ... , v_n \} $$
Since W ⊆ V, the subspace W is included in the vector space V.
Therefore, all vectors {v1,...,vn} of the basis Bw belong to V.
$$ B_W ∈ V $$ $$ \{ v_1 , ... , v_n \} ∈ V $$
This implies that Bw is also a basis B of the vector space W.
Since a vector basis can represent all the vectors of a vector space, the basis Bw can represent both the vectors of the subspace W and those of the vector space V.
This is only possible if W = V.
$$ W = V $$
In conclusion, a subspace W has the same dimension m=n as the vector space V to which it belongs (W ⊆ V), if and only if W=V.
Corollary 3 ( Codimension )
Consider a vector space V over the field K with dimension n and a subspace W ⊆ V of dimension m. The codimension is defined as the difference between the dimension of the vector space V and the dimension of the subspace W. $$ codim_k(W) = dim_k(V) - dim_k(W) = n-m $$
Corollary 4 ( Grassman Theorem )
Given a finite-dimensional vector space V over the field K of dimension n, and two subspaces A and B, the dimension of the sum set dim(A+B) does not equal the sum of the dimensions of the subspaces dim(A)+dim(B). $$ dim_k(A+B) = dim_k(A) + dim_k(B) - dim_k(A∩B) $$
This does not apply if A and B are in direct sum.
Corollary 5
The number of elements in a basis is solely dependent on the vector space. It does not vary depending on the chosen basis.
And so on.
