📂Linear Algebra

Coordinate Transformation of Vectors

Overview^{1 2}

Let $V$ be $n$dimensional vector space, and let us call $\mathbf{v} \in V$ as such. Let $\beta$ be some ordered basis of $V$. Then, $\mathbf{v}$ is expressed as the coordinate vector $[\mathbf{v}]_{\beta}$. Given another ordered basis $\beta ^{\prime}$, $\mathbf{v}$ can also be expressed as the coordinate vector $[\mathbf{v}]_{\beta^{\prime}}$ 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 $V$ as $n = 2$. Let $\beta = \left\{ \mathbf{u}_{1}, \mathbf{u}_{2} \right\}$ be the ordered basis of $V$, and consider another ordered basis $\beta^{\prime} = \left\{ \mathbf{u}_{1}^{\prime}, \mathbf{u}_{2}^{\prime} \right\}$. Suppose the coordinate vectors of $\beta ^{\prime}$ with respect to $\beta$ are given as follows.

$$ [\mathbf{u}_{1}^{\prime}]_{\beta} = \begin{bmatrix} a \\ b \end{bmatrix} \quad \text{and} \quad [\mathbf{u}_{2}^{\prime}]_{\beta} = \begin{bmatrix} c \\ d \end{bmatrix} $$

In other words, the following equation holds.

$$ \begin{equation} \begin{aligned} \mathbf{u}_{1}^{\prime} &= a\mathbf{u}_{1} + b\mathbf{u}_{2} \\ \mathbf{u}_{2}^{\prime} &= c\mathbf{u}_{1} + d\mathbf{u}_{2} \end{aligned} \end{equation} $$

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

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

$$ \mathbf{v} = k_{1}\mathbf{u}_{1}^{\prime} + k_{2}\mathbf{u}_{2}^{\prime} $$

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

$$ \begin{align*} \mathbf{v} &= k_{1}(a\mathbf{u}_{1} + b\mathbf{u}_{2}) + k_{2}(c\mathbf{u}_{1} + d\mathbf{u}_{2}) \\ &= (k_{1}a + k_{2}c)\mathbf{u}_{1} + (k_{1}b + k_{2}d)\mathbf{u}_{2} \end{align*} $$

$$ [\mathbf{v}]_{\beta} = \begin{bmatrix} (k_{1}a + k_{2}c) \\ (k_{1}b + k_{2}d) \end{bmatrix} $$

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

$$ \begin{align*} [\mathbf{v}]_{\beta} &= \begin{bmatrix} (k_{1}a + k_{2}c) \\ (k_{1}b + k_{2}d) \end{bmatrix} \\ &= \begin{bmatrix} a & c \\ b & d \end{bmatrix} \begin{bmatrix} k_{1} \\ k_{2} \end{bmatrix} \\ &= \begin{bmatrix} a & c \\ b & d \end{bmatrix} [\mathbf{v}]_{\beta^{\prime}} \\ \end{align*} $$

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

$$ [\mathbf{v}]_{\beta} = Q[\mathbf{v}]_{\beta^{\prime}} = \begin{bmatrix} [\mathbf{u}_{1}^{\prime}]_{\beta} & [\mathbf{u}_{2}^{\prime}]_{\beta} \end{bmatrix} [\mathbf{v}]_{\beta^{\prime}} \quad \forall \mathbf{v} \in V $$

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

Definition

Let $\beta, \beta^{\prime}$ be two ordered bases of the $n$ dimensional vector space $V$. $Q = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta}$ is called the change of coordinate matrix or transition matrix. For $\mathbf{v} \in V$, the equation below

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

indicates that $Q$ transforms the $\beta^{\prime}$-coordinates into $\beta$-coordinates.

Explanation

Specifically, if we say $\beta = \left\{ \mathbf{u}_{1}, \dots, \mathbf{u}_{n} \right\}$, $\beta^{\prime} = \left\{ \mathbf{u}_{1}^{\prime}, \dots, \mathbf{u}_{n}^{\prime} \right\}$,

$$ Q = \begin{bmatrix} [\mathbf{u}_{1}^{\prime}]_{\beta} & \cdots & [\mathbf{u}_{n}^{\prime}]_{\beta} \end{bmatrix} $$

$$ \mathbf{u}_{j}^{\prime} = \sum_{i} Q_{ij}\mathbf{u}_{i} $$

If $Q$ changes $\beta^{\prime}$-coordinates into $\beta$-coordinates, then $Q^{-1}$ transforms the $\beta$-coordinates into $\beta^{\prime}$-coordinates.

Theorem

Let $\beta, \beta^{\prime}$ be two ordered bases of the $n$ dimensional vector space $V$. Let’s say it’s $Q = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta}$. Then,

(a) $Q$ is an invertible matrix.

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

Proof

(a)

Auxiliary Theorem

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

Since the identity transformation $I$ is invertible, by the auxiliary theorem, $Q$ is invertible.

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(b)

Due to the properties of coordinate vectors and matrix representation,

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

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