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Coordinate Transformation of Vectors 📂Linear Algebra

Coordinate Transformation of Vectors

Overview1 2

Let VV be nndimensional vector space, and let us call v∈V\mathbf{v} \in V as such. Let β\beta be some ordered basis of VV. Then, v\mathbf{v} is expressed as the coordinate vector [v]β[\mathbf{v}]_{\beta}. Given another ordered basis β′\beta ^{\prime}, v\mathbf{v} can also be expressed as the coordinate vector [v]β′[\mathbf{v}]_{\beta^{\prime}} with respect to it. Coordinate transformation of vectors refers to the equation relating these two coordinate vectors.

Build-up

For convenience, let’s call the dimension of VV as n=2n = 2. Let β={u1,u2}\beta = \left\{ \mathbf{u}_{1}, \mathbf{u}_{2} \right\} be the ordered basis of VV, and consider another ordered basis β′={u1′,u2′}\beta^{\prime} = \left\{ \mathbf{u}_{1}^{\prime}, \mathbf{u}_{2}^{\prime} \right\}. Suppose the coordinate vectors of β′\beta ^{\prime} with respect to β\beta are given as follows.

[u1′]β=[ab]and[u2′]β=[cd] [\mathbf{u}_{1}^{\prime}]_{\beta} = \begin{bmatrix} a \\ b \end{bmatrix} \quad \text{and} \quad [\mathbf{u}_{2}^{\prime}]_{\beta} = \begin{bmatrix} c \\ d \end{bmatrix}

In other words, the following equation holds.

u1′=au1+bu2u2′=cu1+du2 \begin{equation} \begin{aligned} \mathbf{u}_{1}^{\prime} &= a\mathbf{u}_{1} + b\mathbf{u}_{2} \\ \mathbf{u}_{2}^{\prime} &= c\mathbf{u}_{1} + d\mathbf{u}_{2} \end{aligned} \end{equation}

Now, choose some vector v∈V\mathbf{v} \in V, and let’s say its coordinate vector with respect to β′\beta^{\prime} is as follows.

[v]β′=[k1k2] \begin{equation} [\mathbf{v}]_{\beta^{\prime}} = \begin{bmatrix} k_{1} \\ k_{2} \end{bmatrix} \end{equation}

v=k1u1′+k2u2′ \mathbf{v} = k_{1}\mathbf{u}_{1}^{\prime} + k_{2}\mathbf{u}_{2}^{\prime}

Upon substituting (1)(1) into the above equation,

v=k1(au1+bu2)+k2(cu1+du2)=(k1a+k2c)u1+(k1b+k2d)u2 \begin{align*} \mathbf{v} &= k_{1}(a\mathbf{u}_{1} + b\mathbf{u}_{2}) + k_{2}(c\mathbf{u}_{1} + d\mathbf{u}_{2}) \\ &= (k_{1}a + k_{2}c)\mathbf{u}_{1} + (k_{1}b + k_{2}d)\mathbf{u}_{2} \end{align*}

[v]β=[(k1a+k2c)(k1b+k2d)] [\mathbf{v}]_{\beta} = \begin{bmatrix} (k_{1}a + k_{2}c) \\ (k_{1}b + k_{2}d) \end{bmatrix}

If we use (2)(2) to simplify the above equation,

[v]β=[(k1a+k2c)(k1b+k2d)]=[acbd][k1k2]=[acbd][v]β′ \begin{align*} [\mathbf{v}]_{\beta} &= \begin{bmatrix} (k_{1}a + k_{2}c) \\ (k_{1}b + k_{2}d) \end{bmatrix} \\ &= \begin{bmatrix} a & c \\ b & d \end{bmatrix} \begin{bmatrix} k_{1} \\ k_{2} \end{bmatrix} \\ &= \begin{bmatrix} a & c \\ b & d \end{bmatrix} [\mathbf{v}]_{\beta^{\prime}} \\ \end{align*}

If we say this is Q=[acbd]Q = \begin{bmatrix} a & c \\ b & d \end{bmatrix}, the coordinate vector with respect to β\beta can be obtained by multiplying the matrix QQ to the coordinate vector with respect to β′\beta^{\prime}. Also, each column of QQ consists of the coordinate vectors of β′\beta^{\prime} with respect to β\beta.

[v]β=Q[v]β′=[[u1′]β[u2′]β][v]β′∀v∈V [\mathbf{v}]_{\beta} = Q[\mathbf{v}]_{\beta^{\prime}} = \begin{bmatrix} [\mathbf{u}_{1}^{\prime}]_{\beta} & [\mathbf{u}_{2}^{\prime}]_{\beta} \end{bmatrix} [\mathbf{v}]_{\beta^{\prime}} \quad \forall \mathbf{v} \in V

Therefore, it becomes evident that Q=[I]β′βQ = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta}. Here, II is an [identity transformation](../3026/#identity transformation).

Definition

Let β,β′\beta, \beta^{\prime} be two ordered bases of the nn dimensional vector space VV. Q=[I]β′βQ = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta} is called the change of coordinate matrix or transition matrix. For v∈V\mathbf{v} \in V, the equation below

[v]β=Q[v]β′ [\mathbf{v}]_{\beta} = Q [\mathbf{v}]_{\beta^{\prime}}

indicates that QQ transforms the β′\beta^{\prime}-coordinates into β\beta-coordinates.

Explanation

Specifically, if we say β={u1,…,un}\beta = \left\{ \mathbf{u}_{1}, \dots, \mathbf{u}_{n} \right\}, β′={u1′,…,un′}\beta^{\prime} = \left\{ \mathbf{u}_{1}^{\prime}, \dots, \mathbf{u}_{n}^{\prime} \right\},

Q=[[u1′]β⋯[un′]β] Q = \begin{bmatrix} [\mathbf{u}_{1}^{\prime}]_{\beta} & \cdots & [\mathbf{u}_{n}^{\prime}]_{\beta} \end{bmatrix}

uj′=∑iQijui \mathbf{u}_{j}^{\prime} = \sum_{i} Q_{ij}\mathbf{u}_{i}

If QQ changes β′\beta^{\prime}-coordinates into β\beta-coordinates, then Q−1Q^{-1} transforms the β\beta-coordinates into β′\beta^{\prime}-coordinates.

Theorem

Let β,β′\beta, \beta^{\prime} be two ordered bases of the nn dimensional vector space VV. Let’s say it’s Q=[I]β′βQ = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta}. Then,

(a) QQ is an invertible matrix.

(b) ∀v∈V\forall \mathbf{v} \in V, [v]β=Q[v]β′[\mathbf{v}]_{\beta} = Q [\mathbf{v}]_{\beta^{\prime}}

Proof

(a)

Auxiliary Theorem

A linear transformation TT being invertible is equivalent to [T]βγ\begin{bmatrix} T \end{bmatrix}_{\beta}^{\gamma} being invertible.

Since the identity transformation II is invertible, by the auxiliary theorem, QQ is invertible.

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(b)

Due to the properties of coordinate vectors and matrix representation,

[v]β=[I(v)]β=[I]β′β[v]β′=Q[v]β′ [\mathbf{v}]_{\beta} = \begin{bmatrix} I(\mathbf{v}) \end{bmatrix}_{\beta} = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta} [\mathbf{v}]_{\beta^{\prime}} = Q [\mathbf{v}]_{\beta^{\prime}}

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  1. Stephen H. Friedberg, Linear Algebra (4th Edition, 2002), p110-112 ↩︎

  2. Howard Anton, Elementary Linear Algebra: Aplications Version (12th Edition, 2019), p256-259 ↩︎