[Linear Algebra] Coordinates and Change of Basis

Coordinates and Change of Bases

Coordinate

All vectors in a vector space is represented as a linear combination of basis of vector space since it spans the space. Then, a natural notation of vectors called coordinate comes up as follows.

$\mathbf{Definition.}$ Coordinates
Assume \(\mathcal{B} = \{ \mathbf{v}_1, \cdots, \mathbf{v}_n \}\) for a vector space $W$ over field $F$. If \(\mathbf{w} = a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n\), then we call $a_1, \cdots, a_n$ be the coordinates of $\mathbf{w}$ with respect to $\mathcal{B}$. We denote

\[\begin{align*} [\mathbf{w}]_\mathcal{B}=\left[\begin{array}{c} a_1 \\ \vdots \\ a_n \end{array}\right] \end{align*}\]

and it is called coordinate matrix.

Here are some examples:

$\mathbf{Example.}$ Standard coordinates

Let \(\mathcal{S} = \{ \mathbf{e}_1, \cdots, \mathbf{e}_n \}\) is the standard basis for $\mathbb{R}^n$. Then, the standard coordinate of the vector $\mathbf{w} = (w_1, \cdots, w_n)$ is

\[\begin{align*} [\mathbf{w}]_\mathcal{S}=\left[\begin{array}{c} w_1 \\ \vdots \\ w_n \end{array}\right] \end{align*}\]

$\mathbf{Example.}$ For orthonormal basis

Let \(\mathcal{B} = \{ \mathbf{v}_1, \cdots, \mathbf{v}_n \}\) is an orthonormal basis for $W$ over $F$. If $\mathbf{w} \in W$, then \(\mathbf{w} = \sum_{i=1}^n \left\langle \mathbf{w}, \mathbf{v}_i \right\rangle \mathbf{v}_i\). Thus

\[\begin{align*} [\mathbf{w}]_\mathcal{B}=\left[\begin{array}{c} \left\langle \mathbf{w}, \mathbf{v}_1 \right\rangle \\ \vdots \\ \left\langle \mathbf{w}, \mathbf{v}_n \right\rangle \end{array}\right] \end{align*}\]

Change of Basis

Note that a basis is not unique for a common vector space $W$. There may be various bases which represents the vector (by a coordinate) much neatle and simple. In that way, it is natural to find the relationship between coordinates with different bases and we state the change of basis problem:

If $\mathbf{w}$ is a vector in vector space $W$ over $F$, and if we change the basis for $W$ from a basis $\mathcal{B}$ to a basis $\mathcal{B}^\prime$, how are the coordinates matrices \([\mathbf{w}]_\mathcal{B}\) and \([\mathbf{w}]_{\mathcal{B}^\prime}\) related?

$\mathbf{Thm\ 1.}$ Solution to the change of basis problem
The coordinate matrices of $\mathbf{w}$ with regard to 2 bases \(\mathcal{B} = \{ \mathbf{v}_1, \cdots, \mathbf{v}_n \}\) and \(\mathcal{B}^\prime= \{ \mathbf{v}_1^\prime, \cdots, \mathbf{v}_n^\prime \}\) are related by the matrix multiplication

\[\begin{align*} [\mathbf{w}]_{\mathcal{B}^\prime} = P_{\mathcal{B} \to \mathcal{B}^\prime} [\mathbf{w}]_\mathcal{B} \end{align*}\]

where

\[\begin{align*} P_{\mathcal{B} \rightarrow \mathcal{B}^{\prime}}=\left[\begin{array}{ll} {\left[\mathbf{v}_1\right]_{\mathcal{B}^{\prime}}} & \cdots\left[\mathbf{v}_n\right]_{\mathcal{B}^{\prime}} \end{array}\right] \end{align*}\]

is called transition matrix from $\mathcal{B}$ to $\mathcal{B}^\prime$, or change of coordinates matrix from $\mathcal{B}$ to $\mathcal{B}^\prime$.

$\mathbf{Proof.}$

Proof is simple. As a basis is also an element of the vector space, we suppose

\[\begin{align*} \mathbf{v}_i = b_{1i} \mathbf{v}_1^\prime + \cdots + b_{ni} \mathbf{v}_n^\prime \end{align*}\]

For $\mathbf{w} \in W$, suppose \(\mathbf{w} = a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n\). Then,

\[\begin{align*} \mathbf{w} &= \sum_{i=1}^n b_{1i} a_i \mathbf{v}_1^\prime + \cdots + \sum_{i=1}^n b_{ni} a_i \mathbf{v}_n^\prime \\ &= a_1^\prime \mathbf{v}_1^\prime + \cdots + a_n^\prime \mathbf{v}_n^\prime \end{align*}\]

Thus, from

\[\begin{align*} \left[\begin{array}{c} a_1^{\prime} \\ \vdots \\ a_k^{\prime} \end{array}\right]=\left[\begin{array}{ccc} b_{11} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nn} \end{array}\right]\left[\begin{array}{c} a_1 \\ \vdots \\ a_n \end{array}\right] \end{align*}\]

Reasonably, all transition matrices are invertible:

$\mathbf{Thm\ 2.}$ Solution to the change of basis problem
If $\mathcal{B}$ and $\mathcal{B}^\prime$ are bases for $W$ over $F$, then the transition matrices \(P_{\mathcal{B}^\prime \to \mathcal{B}}\) and \(P_{\mathcal{B} \to \mathcal{B}^\prime}\) are invertible, and they are inverses of one another, i.e.,

\[\begin{align*} (P_{\mathcal{B}^\prime \to \mathcal{B}})^{-1} = P_{\mathcal{B} \to \mathcal{B}^\prime} \end{align*}\]

$\mathbf{Note.}$ Also, we may expect the transitivity of the transition matrices, and it really does:

\[\begin{align*} P_{\mathcal{B}_2 \to \mathcal{B}_3} P_{\mathcal{B}_1 \to \mathcal{B}_2} = P_{\mathcal{B}_1 \to \mathcal{B}_3} \end{align*}\]

Reference

[1] Contemporary Linear Algebra, Howard Anton, Robert C. Busby