One may get confused when asked to specify the coordinates of T as depicted below.

Figure 1

However, they may think it easy to determine the coordinates of T if instead of Figure 1, they are given the following figure.

Figure 2

As Figure 2 shows, the coordinates of T are (1,1). However, T does not have unique coordinates. In fact, the coordinates (in a two dimensional space) depend on where we put the point of reference O and a pair of nonparallel “yardsticks”. In Figure 3, the coordinates of T are (2,2) because we have used yardsticks whose lengths are half of the previous ones.

Figure 3

Loosely speaking, the coordinates of T are affected not only by the “unit” lengths of the yardsticks but also their directions.

Figure 4

In Figure 4, although the unit lengths of the yardsticks being used are equal, the coordinates of T still depend on whether we use the xy-plane or x’y’-plane. On the xy-plane, the coordinates of T is (2,2), while on the x’y’-plane its coordinates are (2 \sqrt{2},0). In linear algebra, we adopt the term basis (of a vector space) to mean the same thing as yardsticks in the previous discussion. In brief, we claim that the coordinates of T above depend on the basis being used in the \mathbb{R}^2 vector space.

 

Before looking at the definition of the coordinates of a vector \vec{v} in a vector space V, it is worth pointing out that every vector can be expressed as a unique linear combination of some fixed basis vectors, as described more precisely in the following theorem.

 

The Unique Representation Theorem

Let B = \begin{Bmatrix} \vec{b}_1, \vec{b}_2, \cdots \vec{b}_n \end{Bmatrix} be a basis for a vector space V. Then for each \vec{v} in V, there exists a unique set of scalars c1, c2, …, cn such that:

\vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + \cdots + c_n \vec{b}_n

 

Example 1

Consider the vector \vec{v} = \begin{bmatrix}7 \\ 2 \end{bmatrix} \in \mathbb{R}^2 and the basis B = \begin{Bmatrix} \vec{b}_1 , \vec{b}_2 \end{Bmatrix} where \vec{b}_1 = \begin{bmatrix} 2 \\ -1 \end{bmatrix} and \vec{b}_2 = \begin{bmatrix} 1 \\ 5 \end{bmatrix}. Find c1 and c2 such that \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2.

Answer

Let c1 and c2 be real numbers satisfying \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2. Consequently,

\begin{bmatrix}7 \\ 2 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 5 \end{bmatrix} \\ \begin{bmatrix}7 \\ 2 \end{bmatrix} = \begin{bmatrix}2c_1 + c_2 \\ -c_1 + 5c_2 \end{bmatrix}

So we have a system of linear equations with two unknowns c1 and c2. By applying elimination method (or other techniques to solve such problems), it can be shown that c1 = 3 and c2 = 1 satisfy \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2.

 

Example 2

Let Pn be the vector space whose members, or vectors, are all real polynomial functions having degree ≤ n, that is, all functions expressible in the form p(x) = a0 + a1x + … + anxn where a0, a1, …, an are real numbers. A basis for P2 is B = \begin{Bmatrix} \vec{b}_1 , \vec{b}_2 , \vec{b}_3 \end{Bmatrix} where

\\ \vec{b}_1 = 2 + x \\ \vec{b}_2 = -1 + 3x + x^2 \\ \vec{b}_3 = -5x + 2x^2

Given \vec{v} = 6 + 8x - 2x^2, find c1, c2, and c3 such that \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3.

Answer

\vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3

6 + 8x – 2x2 = c1(2 + x) + c2(-1 + 3x + x2) + c3(-5x + 2x2)

6 + 8x – 2x2 = (2c1 – c2) + (c1 + 3c2 – 5c3)x + (c2 + 2c3)x2

From this, we have the following system of linear equations in c1, c2, and c3.

\left\{ \begin{matrix} 2c_1 & - c_2 & & = 6 \\ c_1 & + 3c_2 & - 5c_3 & = 8 \\ & c_2 & +2c_3 & = -2 \end{matrix}

By applying the Cramer’s Rule, for example, it can be shown that c1 = 3, c2 = 0, and c3 = -1 satisfy \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3.

 

Definition

Let B = \begin{Bmatrix} \vec{b}_1 , \vec{b}_2 , \cdots \vec{b}_n \end{Bmatrix} be a basis for a finite dimensional vector space V and \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + \cdots c_n \vec{b}_n be the expression for \vec{v} in terms of the basis B. Then the scalars c1, c2, …, cn are called the coordinates of \vec{v} relative to the basis B. The coordinate vector of \vec{v} relative to B is denoted by (\vec{v})_B and is the vector in \mathbb{R}^n defined by (\vec{v})_B = (c_1,c_2, \cdots , c_n). The coordinate matrix of \vec{v} relative to B is denoted by [ \vec{v} ]_{B} and is the n×1 matrix defined by

\begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}

 

Example 3

Find the coordinate vector and coordinate matrix of \vec{v} relative to the basis B in Example 1.

Answer

It has been shown in Example 1 that c1 = 3 and c2 = 1 satisfy \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2. Thus,

\begin{bmatrix}7 \\ 2 \end{bmatrix} = 3 \begin{bmatrix} 2 \\ -1 \end{bmatrix} + 1 \begin{bmatrix} 1 \\ 5 \end{bmatrix}

By the notation just-defined above, the coordinate vector of \vec{v} relative to B is (\vec{v})_B = (3,1) and its coordinate matrix (relative to B) is

[ \vec{v} ]_{B} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}

 

Example 4

Find the coordinate vector and coordinate matrix of \vec{v} relative to the basis B in Example 2.

Answer

It has been shown in Example 2 that c1 = 3, c2 = 0, and c3 = -1 satisfy \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3. Thus,

6 + 8x – 2x2 = 3(2 + x) + 0(-1 + 3x + x2) + (-1)(-5x + 2x2)

By the notation just-defined above, the coordinate vector of \vec{v} relative to B is (\vec{v})_B = (3,0,-1) and its coordinate matrix (relative to B) is

[ \vec{v} ]_{B} = \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix}

 

Example 5

Referring to Example 2, suppose that we use the basis C = \begin{Bmatrix} \vec{c}_1 , \vec{c}_2 , \vec{c}_3 \end{Bmatrix} instead of B, where

\\ \vec{c}_1 = 1 + 2x - x^2 \\ \vec{c}_2 = 6x - 4x^2 \\ \vec{c}_3 = 5 + 3x^2

Find the coordinate matrix of \vec{v} relative to the “new” basis C.

Answer

By applying the similar method as in Example 2 to find c1, c2, and c3, it can be proved that c1 = 1, c2 = 1, and c3 = 1 satisfy \vec{v} = c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3. Thus,

6 + 8x – 2x2 = 1(1 + 2x – x2) + 1(6x – 4x2) + 1(5 + 3x2)

By the notation just-defined above, the coordinate matrix of \vec{v} relative to C is

[ \vec{v} ]_{C} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

 

It is of interest to know how [ \vec{v} ]_{C} relates to [ \vec{v} ]_{B}. Alternatively, how to express [ \vec{v} ]_{C} in terms of [ \vec{v} ]_{B}? The next theorem answers the question.

 

Theorem

Suppose that B = \begin{Bmatrix} \vec{b}_1 , \vec{b}_2 , \cdots , \vec{b}_n \end{Bmatrix} and C = \begin{Bmatrix} \vec{c}_1 , \vec{c}_2 , \cdots , \vec{c}_n \end{Bmatrix} are bases for a vector space V and \vec{v} ∈ V. Then [ \vec{v} ]_{B} = P [ \vec{v} ]_{C} where the column vectors of P are [ \vec{c}_1 ]_{B}, [ \vec{c}_2 ]_{B}, \cdots [ \vec{c}_n ]_{B}.

Note:

The matrix P above can be denoted by P = \begin{bmatrix} \begin{array}{c:c:c:c} [ \vec{c}_1 ]_{B} & [ \vec{c}_2 ]_{B} & \cdots & [ \vec{c}_n ]_{B} \end{array} \end{bmatrix} and is called the transition matrix from C to B.

 

Example 6

Consider the vector \vec{v} = \begin{bmatrix}-3 \\ 1 \end{bmatrix} \in \mathbb{R}^2 and the following bases for \mathbb{R}^2:

B = \begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{Bmatrix} and C = \begin{Bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ -3 \end{bmatrix} \end{Bmatrix}

  1. Find the coordinate vector and coordinate matrix of \vec{v} relative to B.
  2. By using the transition matrix P as described in the theorem above, determine the coordinate matrix of \vec{v} relative to C, i.e. [ \vec{v} ]_C.
  3. If c1 and c2 are the first and the second coordinates of [ \vec{v} ]_C, respectively, verify that \vec{v} = c_1 \vec{c}_1 + c_2 \vec{c}_2, where \vec{c}_1 = \begin{bmatrix} 1 \\ -2 \end{bmatrix} and \vec{c}_2 = \begin{bmatrix} 2 \\ -3 \end{bmatrix}.

Answer

Part a

By inspection, \begin{bmatrix}-3 \\ 1 \end{bmatrix} = -3 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1 \begin{bmatrix} 0 \\ 1 \end{bmatrix}. Thus the coordinate vector of \vec{v} relative to B is (\vec{v})_B = (-3,1). Consequently, its coordinate matrix is [ \vec{v} ]_B = \begin{bmatrix} -3 \\ 1 \end{bmatrix}.

Part b

Let \vec{c}_1 = \begin{bmatrix} 1 \\ -2 \end{bmatrix} and \vec{c}_2 = \begin{bmatrix} 2 \\ -3 \end{bmatrix}. Since \vec{c}_1 = 1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + (-2) \begin{bmatrix} 0 \\ 1 \end{bmatrix} and \vec{c}_2 = 2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + (-3) \begin{bmatrix} 0 \\ 1 \end{bmatrix}, it follows that the coordinate matrices of \vec{c}_1 and \vec{c}_2 relative to B are [\vec{c}_1]_B = \begin{bmatrix} 1 \\ -2 \end{bmatrix} and [\vec{c}_2]_B = \begin{bmatrix} 2 \\ -3 \end{bmatrix}, respectively. Thus, the transition matrix from C to B is P = \begin{bmatrix} 1 & 2 \\ -2 & -3 \end{bmatrix}. The theorem above states that [ \vec{v} ]_{B} = P [ \vec{v} ]_{C} . Multiplying both sides of the equation by P-1 yields P^{-1} [ \vec{v} ]_{B} = [ \vec{v} ]_{C} . Note that P^{-1} = \begin{bmatrix} -3 & -2 \\ 2 & 1 \end{bmatrix}. So, the coordinate matrix of \vec{v} relative to C can be computed as follows.

[ \vec{v} ]_C = P^{-1} [ \vec{v} ]_B = \begin{bmatrix} -3 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 7 \\ -5 \end{bmatrix}

Part c

Because c1 and c2 are the first and the second coordinates of [ \vec{v} ]_C = \begin{bmatrix} 7 \\ -5 \end{bmatrix}, c1 = 7 and c2 = -5. Consequently,

c_1 \vec{c}_1 + c_2 \vec{c}_2 = 7 \begin{bmatrix} 1 \\ -2 \end{bmatrix} + (-5) \begin{bmatrix} 2 \\ -3 \end{bmatrix} = \begin{bmatrix} -3 \\ 1 \end{bmatrix} = \vec{v}.

So, \vec{v} = c_1 \vec{c}_1 + c_2 \vec{c}_2 is verified.

 

In part b of Example 3, we have used the following corollary to the previous theorem:

Suppose that B = \begin{Bmatrix} \vec{b}_1 , \vec{b}_2 , \cdots , \vec{b}_n \end{Bmatrix} and C = \begin{Bmatrix} \vec{c}_1 , \vec{c}_2 , \cdots , \vec{c}_n \end{Bmatrix} are bases for a vector space V and \vec{v} ∈ V. Then [ \vec{v} ]_{C} = P^{-1} [ \vec{v} ]_{B} where the column vectors of P are [ \vec{c}_1 ]_{B}, [ \vec{c}_2 ]_{B}, \cdots [ \vec{c}_n ]_{B}.

 

Example 7

Referring to Example 2, 4, and 5 above, consider the vector \vec{v} = 6 + 8x - 2x^2 ∈ P2. In Example 4, it is shown that [ \vec{v} ]_{B} = \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix}. Using the transition matrix P as described in the theorem above, determine the coordinate matrix of \vec{v} relative to C, i.e. [ \vec{v} ]_C. Verify that we get the same result as obtained in Example 5.

Answer

The column vectors of P are [ \vec{c}_1 ]_{B}, [ \vec{c}_2 ]_{B}, [ \vec{c}_3 ]_{B}. It can be shown that:

\\ \vec{c}_1 = \frac{5}{12} \vec{b}_1 - \frac{1}{6} \vec{b}_2 - \frac{5}{12} \vec{b}_3 \\ \vec{c}_2 = - \frac{1}{3} \vec{b}_1 - \frac{2}{3} \vec{b}_2 - \frac{5}{3} \vec{b}_3 \\ \vec{c}_3 = \frac{35}{12} \vec{b}_1 + \frac{5}{6} \vec{b}_2 + \frac{13}{12} \vec{b}_3

Therefore,

[\vec{c}_1]_B = \begin{bmatrix} 5/12 \\ -1/6 \\ -5/12 \end{bmatrix}, [\vec{c}_2]_B = \begin{bmatrix} - 1/3 \\ - 2/3 \\ -5/3 \end{bmatrix}, [\vec{c}_3]_B = \begin{bmatrix} 35/12 \\ 5/6 \\ 13/12 \end{bmatrix}

These produce the transition matrix P as follows.

P = \begin{bmatrix} 5/12 & -1/3 & 35/12 \\ -1/6 & -2/3 & 5/6 \\ -5/12 & -5/3 & 13/12 \end{bmatrix}

It can be proved that P^{-1} = \begin{bmatrix} 2 & - \frac{27}{2} & 5 \\ - \frac{1}{2} & 5 & - \frac{5}{2} \\ 0 & \frac{5}{2} & -1 \end{bmatrix}.

To find [ \vec{v} ]_C, apply the corollary above. This yields:

[ \vec{v} ]_C = P^{-1} [ \vec{v} ]_B = \begin{bmatrix} 2 & - \frac{27}{2} & 5 \\ - \frac{1}{2} & 5 & - \frac{5}{2} \\ 0 & \frac{5}{2} & -1 \end{bmatrix} \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

The result is the same as obtained in Example 5.

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