1. If S is a basis of a vector space V then every vector in V has exactly one representation as a linear combination of elements of S.

Proof. 1. Let S be a basis of a vector space V. Then by the definition of bases V=span(S), so every vector in V is equal to a linear combination of vectors from S. It remains to prove that this linear combination is unique. By contradiction, suppose that there exists a vector a in V which is equal to two different linear combinations of vectors from S:

Subtracting the first equality from the second one, we get:

Thus the zero vector is equal to a linear combination of elements from S. Since x_i is not equal to y_i for some i, some of the coefficients of this linear combination are not equal to zero. By the theorem about linearly independent sets of vectors, S is not linearly independent. But this contradicts the assumption that S is a basis. Thus we assumed that a is represented by two different linear combinations of elements of S and deduced a contradiction. This completes the proof of part 1.

2.1 If V has a basis with n elements then every set of vectors in V which has more than n elements is linearly dependent.

Proof. Let {v₁,...,v_n} be a basis of V and let S be a subset of V with m>n elements. We need to prove that S is linearly dependent.

Since {v₁,...,v_n} is a basis of V, every element of S is a linear combination of the v's:

Consider the following n-vectors:

Claim: these vectors are linearly dependent. In order to prove that, by theorem about linear independent sets, we need to find not all zero numbers x₁,..., x_m such that

This leads to a homogeneous system of linear equations with n equations and m unknowns. Since m>n, the number of equations is smaller than the number of unknowns. By the theorem about homogeneous systems of equations, this system has a non-zero solution x₁,...,x_m.

Recall that we have the following equalities:

Multiply the first equality by x₁, the second equality -- by x₂, ..., the last equality by x_m and add them all. It is easy to see that on the left we shall get the vector x₁s₁+...+x_ms_m. On the right we shall get p₁v₁+...+p_nv_n where p_i (i=1,...,n) is the i-th coordinate of the vector x₁t₁+...+x_nt_n. Since this vector is zero, all its coordinates are equal to 0, so the right hand side is equal to the zero vector. Thus x₁s₁+...+x_ms_m=0 and so the zero vector is equal to a linear combination of elements from S not all coefficients of which are 0. This implies, according to the theorem about linearly independent sets that S is linearly dependent.

2.2. If V has a basis with n elements then every set of vectors with fewer than n elements does not span V.

Proof. Suppose that V has basis S with n elements and suppose that T is a subset of V with fewer than n elements. We need to show that T does not span V. By contradiction, assume that T spabs V. By the theorem about throwing away elements, we can find a subset T' of T which is linearly independent and also spans V. Since V=span(T') and T' is linearly independent, T' is a basis of V. Since the number of elements in S is bigger than the number of elements in this basis, we can conclude (using the part 2.1 of our theorem) that S is not linearly independent. But this contradicts the assumption that S is a basis. The proof of the part 2.2 is complete.

3. If V has a basis with n elements then all bases of V have the same number of elements.

Proof follows directly from parts 2.1 and 2.2.