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Any vector in a vector space can be represented uniquely by its coordinate vector.

Let $V$ be a vector space with an ordered basis $\beta $, and let $x$ be any vector in $V$:

We can express $x$ as a linear combination of all the basis vectors in $\beta $:

The coefficients ${a}_{i}$ in the linear combination above play a pivotal role: they specify how much each basis vector affects the result of the linear combination. Therefore, in recognition of that important role we put them into a column vector denoted by ${\left[x\right]}_{\beta}$:

We call the column vector ${\left[x\right]}_{\beta}$ the coordinate vector of $x$ relative to the basis $\beta $. Each element of the coordinate vector specifies how much contribution is made to the linear combination by the corresponding basis vector in the basis $\beta $.

In essence, the coordinate vector tells us which basis vectors make contributions to the linear combination. A zero element in the coordinate vector indicates that the corresponding basis vector does not contribute to the linear combination.

Two examples are provided below in the context of the vector space of polynomials.

Let ${P}_{3}$ be the vector space of the polynomials of degree 3, with the ordered basis $\beta $:

There are four basis vectors in the basis $\beta $ of ${P}_{3}$. All vectors in this space, including the basis vectors themselves, can be uniquely expressed as linear combinations of the four basis vectors.

Let $f\left(x\right)$ be a vector in the vector space ${P}_{3}$:

The coordinate vector of $f\left(x\right)$ is given by${\left[f\left(x\right)\right]}_{\beta}$:

In this example, since all the elements of the coordinate vector are non-zero, all basis vectors contribute to the linear combination.

Let $g\left(x\right)$ be another vector in the vector space ${P}_{3}$:

Then the coordinate vector of $g\left(x\right)$ is given by${\left[g\left(x\right)\right]}_{\beta}$:

In this example, the second and third basis vectors do not make any contribution to the linear combination; only the first and last basis vectors do.

[This content was created with ML, available from the App Store.]