The notion of coordinates

Vector spaces: Coordinates

The notion of coordinates

In Coordinate space, we have seen that, after the selection of a basis, the Euclidean space #\mathbb{E}^3# can be identified with #\mathbb{R}^3#. Here we show that this phenomenon does not stand on its own. In the meantime, we determined that the relevant spaces are in fact vector spaces.

Bases are minimal sets that span a vector space. They have a special property, which we deduce now. If a set of vectors spans a vector space #V#, then any vector of #V# can be written as a linear combination of the set. This expression is not unique in general, but it is if the set is a basis:

Basis criterion

Let #\vec{a}_1,\ldots,\vec{a}_n# be a set of vectors of a vector space #V#, where #n# is a natural number.

The sequence #\vec{a}_1,\ldots,\vec{a}_n# is a basis of #V# if and only if, for each #\vec{x}\in V#, there are unique scalars #x_1,\ldots,x_n# such that
\[
\vec{x}=x_1\cdot \vec{a}_1+\cdots +x_n\cdot \vec{a}_n
\]

Suppose that #\basis{\vec{a}_1,\ldots,\vec{a}_n}# is a basis of #V#. If, in addition to #\vec{x}=x_1\cdot \vec{a}_1+\cdots +x_n\cdot \vec{a}_n #, we also have #\vec{x}=y_1\cdot\vec{a}_1+\cdots + y_n\cdot\vec{a}_n#, then \[
\vec{0}=(x_1 -y_1)\cdot\vec{a}_1 + \cdots + (x_n -y_n ) \cdot\vec{a}_n
\] so #x_1 =y_1,\ldots , x_n =y_n# because of the Dependence criterion. This proves that the writing of #\vec{x}# as a linear combination of #\vec{a}_1,\ldots,\vec{a}_n# is unique if #\basis{\vec{a}_1,\ldots,\vec{a}_n}# is a basis of #V#.

Suppose now that, for every #\vec{x}\in V#, there are unique numbers #x_1,\ldots,x_n# such that
\[
\vec{x}=x_1\cdot \vec{a}_1+\cdots +x_n\cdot \vec{a}_n
\] In that case

each vector #\vec{x}# of #V# lies in #\linspan{\vec{a}_1,\ldots, \vec{a}_n}#, and
the equation \[\vec{0}=x_1\cdot \vec{a}_1+\cdots +x_n\cdot \vec{a}_n
\] has the unique solution #x_1=0\land\cdots\land x_n=0#.

Then, according to the Dependence criterion, #\basis{\vec{a}_1,\ldots,\vec{a}_n}# is independent, and, because #V=\linspan{\vec{a}_1,\ldots, \vec{a}_n}#, is even a basis of #V#. This proves the theorem.

In the vector space #\mathbb{R}#, any number #a# distinct from #0# is a basis. Each number #x# can be written uniquely as #\frac{x}{a}\cdot a#. Thus, the unique coefficient of #x# with respect to the basis #a# is #\frac{x}{a}#.

We prefer to define a basis as a sequence or list of vectors since the order in which they are listed is important: the basis will be used to write down coordinates of arbitrary vectors with respect to the basis, and so a change of order would imply a permutation of the coordinates.

We can represent each vector #\vec{x}# from #V# in a unique way by a row of #n# numbers #\rv{x_1,\ldots,x_n}#. This gives rise to the following definition.

Coordinates
Let #n# be a natural number and #\basis{\vec{a}_1, \ldots ,\vec{a}_n}# a basis of a vector space #V#. If
\[
\vec{x}=x_1\cdot\vec{a}_1+\cdots + x_n\cdot\vec{a}_n
\] then the numbers #x_1,\ldots ,x_n# are called the coordinates of the vector #\vec{x}# with respect to this basis. The vector #\rv{x_1, \ldots , x_n}# is called the coordinate vector of #\vec{x}#. This is itself a vector in #\mathbb{R}^n# if #V# is a real vector space and in #\mathbb{C}^n# if #V# is a complex vector space.

The vector space of all coordinate vectors is called the coordinate space of #V#.

Note: the coordinates of a vector depend on the basis used!

In the vector space #\mathbb{R}#, the number #6# has coordinate #3# with respect to the basis #2# and coordinate #2# with respect to the basis #3#.

Consider the complex #1#-dimensional vector space #\mathbb{C}#. If we restrict the scalars to real numbers (which are, after all, special complex numbers), we can view #\mathbb{C}# as a real vector space. The pair #1#, #\ii# is easily seen to be a basis of this real vector space. Thus, the coordinate space of the complex numbers, viewed as a real vector space, is #\mathbb{R}^2#.

The vectors #\rv{ 1 , 1 } # and #\rv{ 2 , 1 } # form a basis of #\mathbb{R}^2#.

The independence can be derived from the fact that #a\cdot \rv{ 1 , 1 } +b\cdot \rv{ 2 , 1 } =\rv{0,0}# implies #a=b=0#. The third property of the dimension gives that two independent vectors in the two-dimensional space #\mathbb{R}^2# form a basis of this space.

Determine the coordinates of the vector #\rv{-1,4}# with respect to this new basis.

#\rv{9,-5}#

The coordinates of the vector #\rv{-1,4}# with respect to this new basis are the numbers #c# and #d# for which #\rv{-1,4}=c\cdot\rv{ 1 , 1 } +d\cdot\rv{ 2 , 1 } #. Reading out the coordinates yields:
\[\lineqs{ -c-2\cdot d-1&=&0\cr -c-d+4 &=&0\cr} \] This system of equations has the unique solution #c=9# and #d=-5#. Therefore, the coordinate vector of #\rv{-1,4}# with respect to the new basis is #\rv{9,-5}#.

The example shows that it is important to keep track of the basis we work with.

New example