Vector Spaces

4.4 Coordinae Systems

Theorem 7: The Unique Reresentation Theorem

Let $B=\left\{ \textbf{b}_1, \cdots, \textbf{b}_n \right\}$ be a basis for vector space $V$. Then for each $\textbf{x}$ in $V$, there exists a unique set of scalars $c_1, \cdots, c_n$ such that

$$\textbf{x} = c_1 \textbf{b}_1 + \cdots + c_n\textbf{b}_n \tag{1}$$

Proof:

• Since $B$ spans $V$, there exist scalars such that (1) holds.
• Suppose $\textbf{x}$ also has the representation

$$\textbf{x} = d_1 \textbf{b}_1 + \cdots + d_n\textbf{b}_n$$

for scalars $d_1, \cdots, d_n$.

• Then, subtracting, we have

$$\textbf{0} = \textbf{x}- \textbf{x} = (c_1-d_1)\textbf{b}_1 + \cdots +(c_n-d_n)\textbf{b}_n \tag{2}$$

• Since $B$ is linearly independent, the weights in (2) must all be zero. That is, $c_j = d_j$ for $1\le j\le n$.

Definition:

• Suppose $B=\left\{ \textbf{b}_1, \cdots, \textbf{b}_n \right\}$ is a basis for $V$ and $\textbf{x}$ is in $V$.
• The coordinates of $\textbf{x}$ relative to the basis $B$ (or the $B$-coordinate of $\textbf{x}$ ) are the weights $c_1, \cdots, c_n$ such that $\textbf{x} = c_1\textbf{b}_1 + \cdots + c_n \textbf{b}_n$.
• If $c_1, \cdots,c_n$ are the $B$-coordinates of $\textbf{x}$, then the vector in $\mathbb{R}^n$

$$[\textbf{x}]_B = \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}$$

is the coordinate vector of $\textbf{x}$ (relative to $B$), or the **$B$-coordinate vector of $\textbf{x}$.

• The mapping $\textbf{x} \mapsto [\textbf{x}]_B$ is the coordinate mapping (determined by $B$).

• When a basis $B$ for $\mathbb{R}^n$ is fixed, the $B$-coordinate vector of a specified $\textbf{x}$ is easily found, as in the example below.

Example 1:

Let $\textbf{b}_1 = \begin {bmatrix} 1 \\ 0 \end{bmatrix}, \textbf{b}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \textbf{x} = \begin{bmatrix} 1 \\ 6 \end{bmatrix}$, and $B=\left\{ \textbf{b}_1, \textbf{b}_2 \right\}$. Find the coordinate vector $[\textbf{x}]_B$ of $\textbf{x}$ relative to $B$.

Solution:

• The $B$-coordinate $c_1, c_2$ of $\textbf{x}$ satisfy

$$\begin {align} c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} &= \begin{bmatrix} 1 \\ 6 \end{bmatrix} \\ c_1 \textbf{b}_1 + c_2 \textbf{b}_2 &= \textbf{x} \end {align}$$

or

$$\begin {align} \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} &= \begin{bmatrix} 1 \\ 6 \end{bmatrix} \\ \begin{bmatrix}\textbf{b}_1 & \textbf{b}_2 \end{bmatrix} [\textbf{x}]_B &= \textbf{x} \end {align} \tag{3}$$

• This equation can be solved by row operations on an augmented matrix or by using the inverse of the matrix on the left.

• In any case, the solution is $c_1=-2, c_2=3$.

• Thus $\textbf{x}=-2\textbf{b}_1+3\textbf{b}_2$ and

$$[\textbf{x}]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}$$

• See the follwing figure.

• The matrix in (3) changes the $B$-coordinates of a vector $\textbf{x}$ into the standard coordinates for $\textbf{x}$.

• An analogous change of coordinates can be carried out in $\mathbb{R}^n$ for a basis $B=\left\{ \textbf{b}_1, \cdots, \textbf{b}_n \right\}$

• Let $P_B = \begin{bmatrix} \textbf{b}_1 & \textbf{b}_2 & \cdots & \textbf{b}_n \end{bmatrix}$

• Then the vector equation

$$\textbf{x} = c_1 \textbf{b}_1 + c_2 \textbf{b}_2 + \cdots + c_n \textbf{b}_n$$

is equivalent to

$$\textbf{x} = P_B [\textbf{x}]_B \tag{4}$$

• $P_B$ is called the change-of-coordinates matrix from $B$ to the standard basis in $\mathbb{R}^n$.

• Left-multiplication by $P_B$ transforms the coordinate vector $[\textbf{x}]_B$
  into $\textbf{x}$.

• Since the columns of $P_B$ form a basis for $\mathbb{R}^n$, $P_B$ is invertible (by the Invertible Matrix Theorem).

• Left-multiplication by $P_B^{-1}$ converts $\textbf{x}$ into its $B$-coordinate vector:

$$P_B^{-1} \textbf{x} = [\textbf{x}]_B$$

• The correspondence $\textbf{x} \mapsto [\textbf{x}]_B$ produced by $P_B^{-1}$, is the coordinate mapping.

• Since $P_B^{-1}$, is an invertible matrix, the coordinate mapping is a one-to-one linear transformation from $\mathbb{R}^n$ onto $\mathbb{R}^n$, by the Invertible Matrix Theorem.

Theorem 8 :

Let $B=\left\{ \textbf{b}_1, \cdots, \textbf{b}_n \right\}$ be a basis for a vector space $V$. Then the coordinate mapping $\textbf{x} \mapsto [\textbf{x}]_B$, is a one-to-one linear transformation from $V$ onto $\mathbb{R}^n$.

Proof :

• Take two typical vectors in $V$, say,

$$\textbf{u} = c_1 \textbf{b}_1 + \cdots + c_n \textbf{b}_n \\ \textbf{w} = d_1 \textbf{b}_1 + \cdots + d_n \textbf{b}_n$$

• Then, using vector operations,

$$\textbf{u}+\textbf{w} = (c_1+d_1) \textbf{b}_1 + \cdots + (c_n+d_n) \textbf{b}_n$$

• It follows that

$$\begin{align} [ \textbf{u} + \textbf{w} ]_B &=\begin{bmatrix} c_1+d_1 \\ \vdots \\ c_n+d_n \end{bmatrix} \\ &=\begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} + \begin{bmatrix} d_1 \\ \vdots \\ d_n \end{bmatrix} \\ &= [\textbf{u}]_B + [\textbf{w}]_B \end{align}$$

• So the coordinate mapping preserves addition.

• If $r$ is any scalar, then

$$\begin{align} r\textbf{u} &= r (c_1 \textbf{b}_1 + \cdots + c_n \textbf{b}_n) \\ &= (rc_1)\textbf{b}_1 + \cdots + (rc_n) \textbf{b}_n \end{align}$$

• So

$$\begin{align} [r\textbf{u}]_B &= \begin{bmatrix} rc_1 \\ \vdots \\ rc_n \end{bmatrix} \\ &= r \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} \\ &= r [\textbf{u}]_B \end{align}$$

• Thus the coordinate mapping also preserves scalar multiplication and

• hence is a linear transformation.

• one-to-one and onto 증명은 숙제. $\blacksquare$

The linearity of the coordinate mapping can extend to linear combinations.

If $\textbf{u}_1, \cdots, \textbf{u}_p$ are in $V$ and if $c_1, \cdots, c_p$ are scalars, then

$$[c_1 \textbf{u}_1 + \cdots + c_p \textbf{u}_p ] = c_1[\textbf{u}_1]_B +\cdots+ c_p[\textbf{u}_p]_B \tag{5}$$

• In words, (5) says that the $B$-coordinate vector of a linear combination of $\textbf{u}_1, \cdots, \textbf{u}_p$ is the same linear combination of their coordinate vectors.

Isomorphism (동형사상)

The coordinate mapping in Theorem 8 is an important example of an isomorphism from $V$ onto $\mathbb{R}^n$.

• In general, a one-to-one linear transformation from a vector space $V$ onto a vector space $W$ is called an isomorphism from $V$ onto $W$.
• The notation and terminology for $V$ and $W$ may differ, but the two spaces are indistinguishable as vector spaces.
• Every vector space calculation in $V$ is accurately reproduced in $W$, and vice versa.
• In particular, any real vector space with a basis of $n$ vectors is indistinguishable from $\mathbb{R}^n$.

Example 5:

Example 7:

Let $\textbf{v}_1=\begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix}, \textbf{v}_2=\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \textbf{x}=\begin{bmatrix} 3 \\ 12 \\ 7 \end{bmatrix}$, and $B=\left\{ \textbf{v}_1, \textbf{v}_2 \right\}$.

Then $B$ is a basis for $H= \text{Span }\left\{ \textbf{v}_1, \textbf{v}_2 \right\}$.

Determine if $\bf{x}$ is in $H$, and if it is, find the coordinate vector of $\bf{x}$ relative to $B$.

Solution :

• If $\textbf{x}$ is in $H$, then the following vector equation is consistent:

$$c_1\begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix} + c_2\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 12 \\ 7 \end{bmatrix}$$

• The scalars $c_1$ and $c_2$, if they exist, are the $B$- coordinates of $\textbf{x}$.

• Using row operations, we obtain

$$\begin{bmatrix} 3 & -1 & 3 \\ 6 & 0 & 12 \\ 2 & -1 & 7 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix}$$

• Thus $c_1=2, c_2=3$ and $[\textbf{x}]_B = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$.

• The coordinate system on $H$ determined by $B$ is shown in the following figure.