The Coordinates (or Weights) of a Vector
The coordinates (or weights) of a vector v relative to a basis B are the scalar coefficients a1,...,an of the linear combination that uniquely determines the vector in the vector space.
The Definition
Given a basis B consisting of a finite number of elements from the vector space V in the field K, for every vector in v, there exists a unique set of scalar coefficients a1,...,an (referred to as coordinates or weights) such that v = a1 v1 + ... + an vn.
The scalars a1,...,an are referred to as coordinates only when a vector is defined by a unique linear combination.
Hence, once a basis is chosen, any vector is defined by a unique linear combination of scalar numbers (coordinates).
v = a1 v1 + ... + an vn
Examples and Exercises on Vector Coordinates
Example 1
Consider a vector space V = R2 in the field R, where every vector can be represented using the canonical basis.
$$ v_1 = (1,0) $$ $$ v_2 = (0,1) $$
Therefore
$$ B = \{ v_1 = (1,0), v_2 = (0,1) \} $$
Let's represent an arbitrary vector of the vector space, for instance, vector v = (4, 2).
The linear combination of vector v is as follows:
$$ (x,y) = a_1 v_1 + a_2 v_2 $$
$$ (4,2) = a_1 (1,0) + a_2 (0,1) $$
$$ (4,2) = (a_1,0) + (0,a_2) $$
$$ (4,2) = (a_1,a_2) $$
which means
$$ \begin{cases} a_1 = 4 \\ a_2 = 2 \end{cases} $$
The coordinates of vector v = (4,2) using the canonical basis are a1=4 and a2=2.
Verification
$$ (x,y) = a_1 v_1 + a_2 v_2 $$
$$ (4,2) = a_1 (1,0) + a_2 (0,1) $$
$$ (4,2) = 4 (1,0) + 2 (0,1) $$
$$ (4,2) = (4,0) + (0,2) $$
$$ (4,2) = (4+0,0+2) = (4,2) $$
Note. These coordinates are unique. Thus, vector v can only be represented using the basis B with the coefficients a1=4 and a2=2.
Example 2
Now, let's represent the same vector v=(4,2) using a different basis B2 in the vector space V.
$$ B_2 = \{ v_1 = (1,1), v_2 = (2,1) \} $$
The linear combination of vector v=(4,2) with the basis B2 is as follows:
$$ (x,y) = a_1 v_1 + a_2 v_2 $$
$$ (4,2) = a_1 (1,1) + a_2 (2,1) $$
$$ (4,2) = (a_1,a_1) + (2a_2,a_2) $$
$$ (4,2) = (a_1 + 2a_2,a_1 + a_2) $$
which means
$$ \begin{cases} a_1 + 2a_2 = 4 \\ a_1 + a_2 = 2 \end{cases} $$
To find the coordinates of the vector, we must solve the system.
$$ \begin{cases} a_1 + 2a_2 = 4 \\ a_1 = 2 - a_2 \end{cases} $$
$$ \begin{cases} (2 - a_2) + 2a_2 = 4 \\ a_1 = 2 - a_2 \end{cases} $$
$$ \begin{cases} a_2 = 4-2 = 2 \\ a_1 = 2 - a_2 \end{cases} $$
$$ \begin{cases} a_2 = 2 \\ a_1 = 2 - (2) = 0 \end{cases} $$
$$ \begin{cases} a_2 = 2 \\ a_1 = 0 \end{cases} $$
The coordinates of vector v = (4,2) using the basis B2 are a1=0 and a2=2.
Verification
$$ (x,y) = a_1 v_1 + a_2 v_2 $$
$$ (4,2) = a_1 (1,1) + a_2 (2,1) $$
$$ (4,2) = 0 (1,1) + 2 (2,1) $$
$$ (4,2) = (0,0) + (4,2) $$
$$ (4,2) = (0+4,0+2) = (4,2) $$
Note. These coordinates are unique. Hence, vector v can be represented using the basis B2 only with the coefficients a1=0 and a2=2.
In conclusion, to represent the same vector v=(4,2) with the basis B2, different scalar coefficients (coordinates) are used compared to the canonical basis or any other basis of the vector space.
And so on.
