Ordered Bases

Let ${\cal B}=\{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n \}$ be a basis of a vector space $V ({\mathbb{F}}).$ As ${\cal B}$ is a set, there is no ordering of its elements. In this section, we want to associate an order among the vectors in any basis of $V.$

DEFINITION 3.4.1 (Ordered Basis)   An ordered basis for a vector space $V ({\mathbb{F}})$ of dimension $n,$ is a basis $\{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n\}$ together with a one-to-one correspondence between the sets $\{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n\}$ and $\{1,2,3, \ldots, n\}.$

If the ordered basis has ${\mathbf u}_1$ as the first vector, ${\mathbf u}_2$ as the second vector and so on, then we denote this ordered basis by

$$({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n).$$

EXAMPLE 3.4.2   Consider ${\cal P}_2({\mathbb{R}}),$ the vector space of all polynomials of degree less than or equal to $2$ with coefficients from ${\mathbb{R}}.$ The set $\{1-x, 1+x, x^2\}$ is a basis of ${\cal P}_2({\mathbb{R}}).$

For any element $a_0 + a_1 x + a_2 x^2 \in {\cal P}_2({\mathbb{R}}),$ we have

$$a_0 + a_1 x + a_2 x^2 = \frac{a_0 - a_1}{2} (1-x) + \frac{a_0 + a_1}{2} (1+x) + a_2 x^2.$$

If $(1-x, 1+x, x^2)$ is an ordered basis, then $$\frac{a_0 - a_1}{2}$$ is the first component, $$\frac{a_0 + a_1}{2}$$ is the second component, and $a_2$ is the third component of the vector $a_0 + a_1 x + a_2 x^2.$

If we take $(1+x, 1-x, x^2)$ as an ordered basis, then $$\frac{a_0 + a_1}{2}$$ is the first component, $$\frac{a_0 - a_1}{2}$$ is the second component, and $a_2$ is the third component of the vector $a_0 + a_1 x + a_2 x^2.$

That is, as ordered bases $({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n),$ $({\mathbf u}_2, {\mathbf u}_3, \ldots, {\mathbf u}_n, {\mathbf u}_1),$ and $({\mathbf u}_n, {\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_{n-1})$ are different even though they have the same set of vectors as elements.

DEFINITION 3.4.3 (Coordinates of a Vector)   Let ${\cal B}= ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n)$ be an ordered basis of a vector space $V ({\mathbb{F}})$ and let ${\mathbf v}\in V.$ If

$${\mathbf v}= \beta_1 {\mathbf v}_1 + \beta_2 {\mathbf v}_2 + \cdots + \beta_n {\mathbf v}_n$$

then the tuple $(\beta_1, \beta_2, \ldots, \beta_n)$ is called the coordinate of the vector ${\mathbf v}$ with respect to the ordered basis ${\cal B}.$

Mathematically, we denote it by $[{\mathbf v}]_{{\cal B}} = (\beta_1, \ldots, \beta_n)^t,$ A COLUMN VECTOR.

Suppose ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n)$ and ${\cal B}_2= ({\mathbf u}_n, {\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_{n-1})$ are two ordered bases of $V.$ Then for any ${\mathbf x}\in V$ there exists unique scalars $\alpha_1, \alpha_2,\dots, \alpha_n$ such that

$${\mathbf x}= \alpha_1 {\mathbf u}_1 + \alpha_2 {\mathbf u}_2 + \c... ...mathbf u}_n + \alpha_1 {\mathbf u}_1 + \cdots + \alpha_{n-1} {\mathbf u}_{n-1}.$$

Therefore,

$$[{\mathbf x}]_{{\cal B}_1} = ({\alpha}_1, {\alpha}_2, \ldots, {\a... ...thbf x}]_{{\cal B}_2} = (\alpha_n, \alpha_1, \alpha_2, \ldots, \alpha_{n-1})^t.$$

Note that ${\mathbf x}$ is uniquely written as $\sum\limits_{i=1}^n \alpha_i {\mathbf u}_i$ and hence the coordinates with respect to an ordered basis are unique.

Suppose that the ordered basis ${\cal B}_1$ is changed to the ordered basis ${\cal B}_3 = ({\mathbf u}_2, {\mathbf u}_1, {\mathbf u}_3, \ldots, {\mathbf u}_n).$ Then $[{\mathbf x}]_{{\cal B}_3} = (\alpha_2, \alpha_1, \alpha_3, \ldots, \alpha_n )^t.$ So, the coordinates of a vector depend on the ordered basis chosen.

EXAMPLE 3.4.4   Let $V = {\mathbb{R}}^3.$ Consider the ordered bases
${\cal B}_1=\bigl( (1,0,0), (0,1,0), (0,0,1) \bigr),$ ${\cal B}_2= \bigl( (1,0,0), (1,1,0), (1,1,1) \bigr)$ and ${\cal B}_3 = \bigl( (1,1,1), (1,1,0), (1,0,0)\bigr)$ of $V.$ Then, with respect to the above bases we have

$$(1, -1, 1) = 1 \cdot (1,0,0) + (-1)\cdot (0, 1, 0) + 1 \cdot (0,0,1).$$

$$= 2 \cdot (1,0,0) + (-2) \cdot (1,1,0) + 1 \cdot (1,1,1).$$

$$= 1 \cdot (1,1,1) + (-2) \cdot (1,1,0) + 2 \cdot (1,0,0).$$

Therefore, if we write ${\mathbf u}= (1,-1,1),$ then

$$[{\mathbf u}]_{{\cal B}_1} = (1,-1,1)^t, \; [{\mathbf u}]_{{\cal B}_2} = (2,-2,1)^t, \; [{\mathbf u}]_{{\cal B}_3} = (1,-2,2)^t.$$

In general, let $V$ be an $n$ -dimensional vector space with ordered bases ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n)$ and ${\cal B}_2 = ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n).$ Since, ${\cal B}_1$ is a basis of $V,$ there exists unique scalars $a_{ij}, \; 1 \leq i, j \leq n$ such that

$${\mathbf v}_i = \sum\limits_{l=1}^n a_{li} {\mathbf u}_l \hspace{0.5in} {\mbox{for }} 1 \leq i \leq n.$$

That is, for each $i, \; 1 \leq i \leq n, \; [{\mathbf v}_i]_{{\cal B}_1} = (a_{1i}, a_{2i}, \ldots, a_{ni})^t.$

Let ${\mathbf v}\in V$ with $[{\mathbf v}]_{{\cal B}_2} = (\alpha_1, \alpha_2, \ldots, \alpha_n)^t.$ As ${\cal B}_2$ as ordered basis $({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n),$ we have

$${\mathbf v}= \sum_{i=1}^n \alpha_i {\mathbf v}_i = \sum_{i=1}^n \... ...ight) = \sum_{j=1}^n \left( \sum_{i=1}^n a_{ji} \alpha_i \right) {\mathbf u}_j.$$

Since ${\cal B}_1$ is a basis this representation of ${\mathbf v}$ in terms of ${\mathbf u}_i$ 's is unique. So,

$$[{\mathbf v}]_{{\cal B}_1} = \left( \sum_{i=1}^n a_{1i} \alpha_i, \sum_{i=1}^n a_{2i} \alpha_i, \ldots, \sum_{i=1}^n a_{ni} \alpha_i \right)^t$$

$$= \begin{bmatrix}a_{11} & \cdots & a_{1n} \\ a_{21} &\cdots & a_{2... ...matrix} \begin{bmatrix} \alpha_1\\ \alpha_2 \\ \vdots \\ \alpha_n \end{bmatrix}$$

$$= A [{\mathbf v}]_{{\cal B}_2}.$$

Note that the $i^{th}$ column of the matrix $A$ is equal to $[{\mathbf v}_i]_{{\cal B}_1},$ i.e., the $i^{th}$ column of $A$ is the coordinate of the $i^{th}$ vector ${\mathbf v}_i$ of ${\cal B}_2$ with respect to the ordered basis ${\cal B}_1.$ Hence, we have proved the following theorem.

THEOREM 3.4.5   Let $V$ be an $n$ -dimensional vector space with ordered bases ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n)$ and ${\cal B}_2 = ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n).$ Let

$$A = \left[ [{\mathbf v}_1]_{{\cal B}_1}, [{\mathbf v}_2]_{{\cal B}_1}, \ldots, [{\mathbf v}_n]_{{\cal B}_1} \right].$$

Then for any ${\mathbf v}\in V,$

$$[v]_{{\cal B}_1} = A [{\mathbf v}]_{{\cal B}_2}.$$

EXAMPLE 3.4.6   Consider two bases ${\cal B}_1= \bigl((1,0,0), (1,1,0), (1,1,1)\bigr)$ and ${\cal B}_2 = \bigl((1,1,1), (1,-1,1), (1,1,0)\bigr)$ of ${\mathbb{R}}^3.$

1. Then
   

   $$[(x,y,z)]_{{\cal B}_1} = (x-y) \cdot (1,0,0) + (y-z) \cdot (1, 1, 0) + z \cdot (1,1,1)$$

   $$= (x-y,y-z,z)^t$$

   
   and
   

   $$[(x,y,z)]_{{\cal B}_2} = (\frac{y-x}{2} + z) \cdot (1,1,1) + \frac{x-y}{2} \cdot (1, -1, 1)$$

   $$+ (x-z) \cdot (1,1,0)$$

   $$= (\frac{y-x}{2} + z, \frac{x-y}{2}, x-z)^t.$$

   
   
2. Let $A = [a_{ij}] = \begin{bmatrix}0 & 2 & 0\\ 0 & -2 & 1 \\ 1 & 1 & 0 \end{bmatrix}.$ The columns of the matrix $A$ are obtained by the following rule:
   $$[(1,1,1)]_{{\cal B}_1} = 0 \cdot (1,0,0) + 0 \cdot (1,1,0) + 1 \cdot (1,1,1) = (0,0,1)^t,$$
   $$[(1,-1,1)]_{{\cal B}_1} = 2 \cdot (1,0,0) + (-2) \cdot (1,1,0) + 1 \cdot (1,1,1) = (2, -2, 1)^t$$
   and
   $$[(1,1,0)]_{{\cal B}_1} = 0 \cdot (1,0,0) + 1 \cdot (1,1,0) + 0 \cdot (1,1,1) = (0,1,0)^t.$$
   That is, the elements of ${\cal B}_2 = \bigl((1,1,1), (1,-1,1), (1,1,0)\bigr)$ are expressed in terms of the ordered basis ${\cal B}_1.$
3. Note that for any $(x,y,z)\in {\mathbb{R}}^3,$
   $$[(x,y,z)]_{{\cal B}_1} = \begin{bmatrix}x-y\\ y-z\\ z \end{bmatri... ...-x}{2} + z\\ \frac{x-y}{2}\\ x-z \end{bmatrix} =A \; \; [(x,y,z)]_{{\cal B}_2}.$$

4. The matrix $A$ is invertible and hence $[(x,y,z)]_{{\cal B}_2} = A^{-1} \;\; [(x,y,z)]_{{\cal B}_1}.$

In the next chapter, we try to understand Theorem 3.4.5 again using the ideas of `linear transformations / functions'.

EXERCISE 3.4.7  

1. Determine the coordinates of the vectors $(1,2,1)$ and $(4, -2, 2)$ with respect to the basis ${\cal B}= \bigl( (2,1,0), (2,1,1), (2,2,1) \bigr)$ of ${\mathbb{R}}^3.$
2. Consider the vector space ${\cal P}_3({\mathbb{R}}).$
   1. Show that ${\cal B}_1= (1 - x, 1 + x^2, 1 - x^3, 3 + x^2 - x^3)$ and ${\cal B}_2= (1, 1- x, 1 + x^2, 1 - x^3)$ are bases of ${\cal P}_3({\mathbb{R}}).$
   2. Find the coordinates of the vector ${\mathbf u}=1 + x + x^2 + x^3$ with respect to the ordered basis ${\cal B}_1$ and ${\cal B}_2.$
   3. Find the matrix $A$ such that $[{\mathbf u}]_{{\cal B}_2} = A [{\mathbf u}]_{{\cal B}_1}.$
   4. Let ${\mathbf v}= a_0 + a_1 x + a_2 x^2 + a_3 x^3.$ Then verify the following:
      

      $$[{\mathbf v}]_{{\cal B}_1} = \begin{bmatrix}-a_1 \\ -a_0 - a_1 + 2 a_2 - a_ 3 \\ -a_0 - a_1 + a_2 - 2 a_3 \\ a_0 + a_1 - a_2 + a_3 \end{bmatrix}$$

      $$= \begin{bmatrix}0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 ... ...\;\; \begin{bmatrix}a_0 + a_1 - a_2 + a_3 \\ -a_1 \\ a_2 \\ -a_3 \end{bmatrix}$$

      $$= [{\mathbf v}]_{{\cal B}_2}.$$