Uniqueness Theorem for Vector Representation in a Basis
Every vector in a vector space is represented by a vector basis B through a unique linear combination of scalars.
The Definition
Let B={v1,v2,...,vn} be a basis of the vector space V over the field K, then every vector v in V can be expressed as a unique linear combination $$ v = \alpha_1 v_1 + \alpha_2 v_2 + ... + \alpha_n v_n $$ with a finite number of coefficients α1,...αn ∈ K
Proof
Consider a generic vector v in the vector space V
$$ \vec{v} \in V $$
The basis of the vector space consists of n vectors {v1,v2,...,vn}
$$ B = \{ \vec{v}_1 , \vec{v}_2 , ... , \vec{v}_n \} $$
Hence, the vector v can be written as a linear combination of the n basis vectors
$$ \vec{v} = \alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 +... + \alpha_n \vec{v}_n $$
For the sake of contradiction, assume that the vector can also be written as another linear combination of the vectors {v1,v2,...,vn}
$$ \vec{v} = \beta_1 \vec{v}_1 + \beta_2 \vec{v}_2 +... + \beta_n \vec{v}_n $$
As it is the same vector, the two linear combinations must be equal
$$ \vec{v} = \vec{v} $$
$$ \alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 +... + \alpha_n \vec{v}_n = \beta_1 \vec{v}_1 + \beta_2 \vec{v}_2 +... + \beta_n \vec{v}_n $$
Move all terms to one side of the equation
$$ \alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 +... + \alpha_n \vec{v}_n - \beta_1 \vec{v}_1 - \beta_2 \vec{v}_2 ... - \beta_n \vec{v}_n = 0 $$
Then, factor out the common terms
$$ ( \alpha_1 - \beta_1 ) \cdot \vec{v}_1 + ( \alpha_2 - \beta_2 ) \cdot \vec{v}_2 +... + ( \alpha_n - \beta_n ) \cdot \vec{v}_n = 0 $$
Since the set of vectors {v1,v2,...,vn} forms a basis of the vector space, by the definition of a basis, the vectors are linearly independent.
When vectors are linearly independent, their linear combination equals zero only in the trivial case, namely when all coefficients are zero.
$$ \underbrace{ ( \alpha_1 - \beta_1 ) }_0 \cdot \vec{v}_1 + \underbrace{ ( \alpha_2 - \beta_2 ) }_0 \cdot \vec{v}_2 +... + \underbrace{( \alpha_n - \beta_n ) }_0 \cdot \vec{v}_n = 0 $$
Therefore, the differences in coefficients are zero
$$ \alpha_1 - \beta_1 = 0 \\ \alpha_2 - \beta_2 = 0 \\ \vdots \\ \alpha_n - \beta_n = 0 $$
This implies that the coefficients are equal
$$ \alpha_1 = \beta_1 \\ \alpha_2 = \beta_2 \\ \vdots \\ \alpha_n = \beta_n $$
The two linear combinations are, in fact, the same.
This proves that the linear combination representing a vector through a basis of the vector space is unique.
Alternate Proof
A basis is a minimal set of vector generators.
Thus, any vector in the vector space can be represented by a linear combination of the n vectors of the basis using n scalars.
$$ v = a_1 v_1 + ... + a_n v_n $$
I need to prove that this representation of a vector using the basis is the only possible one.
According to the theorem, every vector in the space has a unique representation in coordinates relative to the basis, meaning there is a unique n-tuple of scalars a1,...,an such that v=a1v1+...+anvn.
To prove this, I consider the contrary hypothesis.
If two distinct linear combinations (v and w) determined the same vector (v) in the vector space
$$ v = a_1 v_1 +...+ a_n v_n $$ $$ v = b_1 w_1 +...+ b_m w_m $$
this leads to the following equation:
$$ a_1 v_1 +...+ a_n v_n = b_1 w_1 +...+ b_m w_m $$
$$ a_1 v_1 +...+ a_n v_n - b_1 w_1 - ... - b_m w_m = 0 $$
Take the smaller value between n and m, i.e., the lesser number of elements in the two linear combinations
$$ k = min(n,m) $$
Now suppose that up to the k-th element, vectors v and w are the same, while from the k+1-th element onwards, they differ.
$$ v_1 = w_1 ... v_k = w_k $$
$$ v_{k+1} \ne w_{k+1} ... v_n \ne w_m $$
So, I can rewrite the equation as follows:
$$ v_1 ( a_1 - b_1 ) +...+ v_k ( a_k - b_k ) +...+ a_{k+1} v_{k+1} - b_{k+1} w_{k+1} +...+ a_n v_n - b_m w_n = 0 $$
The equation equals zero only if:
$$ a_1=b_1 ... a_k=b_k $$
$$ a_{k+1}=0 ... a_n=0 $$
$$ b_{k+1}=0 ... b_n=0 $$
Up to the k-th element, vectors v and w share the same components, so the only way to nullify the sum is when the scalars a and b are equal (a1=b1 ... ak=bk).
$$ v_1 ( 0 ) +...+ v_k ( 0 ) +...+ a_{k+1} v_{k+1} - b_{k+1} w_{k+1} +...+ a_n v_n - b_m w_n = 0 $$
From the k+1-th element onwards, vectors v and w differ (v≠w). Being linearly independent, the only way to nullify the sum is by setting the scalars a and b to zero (ak+1=0,...,an=0 and bk+1=0,...,bm=0).
$$ v_1 ( 0 ) +...+ v_k ( 0 ) +...+ 0 v_{k+1} - 0 w_{k+1} +...+ 0 v_n - 0 w_n = 0 $$
Only in this way is the equation equal to zero and true.
However, this is a trivial linear combination.
Therefore, to represent the vector v through the basis B, one must necessarily use a unique combination of k scalars.
$$ a_1 = b_1 , ..., a_k = b _k $$
These coefficients are known as coordinates or weights of the vector with respect to the basis.
In conclusion, if the vectors of the basis are linearly independent, there exists a unique n-tuple of scalars in the representation of every vector in the vector space.
