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Categories: Linear algebra

Coordinates vector

Coordinates vector is a very important concept in Linear algebra and representation theory.

Contents

1 Definition

1.1 Example 1
1.2 Example 2

2 Basis transformation matrix

Definition

Let V be a linear space with dim V = n and let

$B = \{ b_1, b_2, b_3, \cdots, b_n \}$

be a linear basis for V. Therefore for every $v \in V$ there is a linear combination (which is unique to v) of the basis vectors such as

$v = \alpha _1 b_1 + \alpha _2 b_2 + \cdots + \alpha _n b_n$

The α-s are determined uniquely by v and B (the theorem of basis guarantees this) and therefore we can say that the following is a Representation of v in the B basis. Now, we define the coordinates vector of v according to B (also called B representation of v) by:

$[ v ]_B = \begin{bmatrix} \alpha _1 \\ \ddots \\ \alpha _n \end{bmatrix}$

and the α-s are called the coordinates of v.

The mapping which matches each vector v from V to its coordinate vector [v]_B is an isomorphism: a linear transformation which is a one-to-one correspondence and onto. This means that every finite-dimensional linear space can be treated as a "columns and squares" space Fⁿ where n is the dimension of V and F is the field on which V is defined.

Coordinates vector is a very important concept in Linear algebra and representation theory, since it allows every calculation with abstract objects to be transformed into a calculation with blocks of numbers (matrices, column vectors) which we know how to do explicitly.

Example 1

Let P3 be the space of all the algebraic polynomials in degree less than 4 (i.e. the highest exponent of x can be 3). This space is linear and spanned by the following polynomials:

B P = {1, x, x 2, x 3}

matching

$1 := \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \quad ; \quad x := \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \quad ; \quad x^2 := \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \quad ; \quad x^3 := \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \quad$

then the corresponding coordinate vector to the polynomial

$p \left( x \right) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ is $\begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix}$ .

According to that representation, the differentiation operator d/dx which we shall mark D will be represented by the following matrix:

$Dp(x) = P'(x) \quad ; \quad [D] = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$

Using that method it is easy to explore the properties of the operator: such as invertability, hermitian or anti-hermitian or none, spectrum and eigenvalues and more.

Example 2

The Pauli matrices which represent the spin operator when transforming the spin eigenstates into vector coordinates.

Basis transformation matrix

Let's mark with [M]^B the matrix which has columns consisting of b₁, b₂, ..., b_n . Then,

v = [M] B [v] B

This formalism can be generalized for transforming v from B representation to a C representation (where C is another basis). Defining basis transformation matrix from B to C as the following matrix:

$[M]_{C}^{B} = \begin{bmatrix} \ [b_1]_C & \cdots & [b_n]_C \ \end{bmatrix}$

we receive the following theorem:

$[v]_C = [M]_{C}^{B} [v]_B$

Corollary:
This matrix is Invertible matrix and M^-1 is the basis transformation matrix from C to B. In other words,

$[M]_{C}^{B} [M]_{B}^{C} = [M]_{C}^{C} = Id$

$[M]_{B}^{C} [M]_{C}^{B} = [M]_{B}^{B} = Id$

Remarks:

The basis transformation matrix can be regarded as an automorphism over V.
$[M]_{E}^{B} = [M]^{B}$ where E is the standard basis.
In order to easily remember the theorem

$[v]_C = [M]_{C}^{B} [v]_B$

notice that the M's sup-index and v's sub-index are "canceling" each other and the M's sub-index is what remains and become v's new sub-index. The "canceling" of index is not a real canceling but rather a manipulation of symbols which serves us for purposes of convenience.

Categories: Linear algebra

Science Fair Project Encyclopedia Contents Page

10-26-2009 08:16:03
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