Exercises: Change of Coordinates

Problem 1

You are given the ordered basis \(\EuScript{B}= (\Vect{b}_1,\Vect{b}_2)\) of \(\RNrSpc{2}\), where

\[\Vect{b}_1 = (1,3) \quad\text{and}\quad \Vect{b}_2 = (-2,-1)\]

Find the coordinate conversion matrix \(\Mtrx{C}_{\EuScript{B}\EuScript{S}}\) from standard \(\EuScript{S}\)-coordinates to \(\EuScript{B}\)-coordinates.
- Solution Answer
Find \(\Vect{x}_{\EuScript{B}}\) if \(\Vect{x}_{\EuScript{S}} = (5,-4)\).
- Answer

Problem 2

You are given two ordered bases of \(\RNrSpc{3}\): \(\EuScript{B}=(\Vect{b}_1,\Vect{b}_2,\Vect{b}_3)\) and \(\EuScript{C}=(\Vect{c}_1,\Vect{c}_2,\Vect{c}_3)\) with

\[ \begin{array}{cclcclccl} \Vect{b}_1 & = & (1,0,1) & \Vect{b}_2 & = & (-1,-1,0) & \Vect{b}_3 & = & (0,-1,-1) \\ \Vect{c}_1 & = & (1,0,0) & \Vect{c}_2 & = & (-1,-1,1) & \Vect{c}_3 & = & (0,0,2) \end{array} \]

Find the coordinate conversion matrix \(\Mtrx{C}_{\EuScript{C}\EuScript{B}}\) from \(\EuScript{B}\)-coordinates to \(\EuScript{C}\)-coordinates.
- Solution Answer
If \(\Vect{x}_{\EuScript{B}} = (2,0,-1)\), find \(\Vect{x}_{\EuScript{C}}\).
- Hint Answer

Problem 3

You are given two ordered bases \(\EuScript{B}=(\Vect{b}_1,\dots ,\Vect{b}_n)\) and \(\EuScript{C}=(\Vect{c}_1,\dots ,\Vect{c}_n)\) of \(\RNrSpc{n}\).

Show that the RREF of the \((n,2n)\)-matrix \([\Mtrx{C}_{\EuScript{S}\EuScript{C}}\ |\ \Mtrx{A}_{\EuScript{S}\EuScript{B}}]\) is the \((n,2n)\)-matrix \([\IdMtrx{n}\ |\ \Mtrx{C}_{\EuScript{C}\EuScript{B}}]\).
Find \(\Mtrx{C}_{\EuScript{B}\EuScript{C}}\), the coordinate conversion matrix from \(\EuScript{C}\)-coordinates to \(\EuScript{B}\)-coordinates, given the bases
\[\EuScript{B}=(\Vect{b}_1,\Vect{b}_2,\Vect{b}_3,\Vect{b}_4) \quad\text{and}\quad \EuScript{C}=(\Vect{c}_1,\Vect{c}_2,\Vect{c}_3,\Vect{c}_4)\]
of \(\RNrSpc{4}\), where
\[ \begin{array}{cclcclcclccl} \Vect{b}_1 & = & (1,1,1,1) & \Vect{b}_2 & = & (0,-1,0,1) \\ \Vect{b}_3 & = & (-1,0,-1,0) & \Vect{b}_4 & = & (0,0,1,-1) \\ \Vect{c}_1 & = & (1,-1,1,2) & \Vect{c}_2 & = & (4,1,3,1) \\ \Vect{c}_3 & = & (-1,2,-1,1) & \Vect{c}_4 & = & (1,2,3,4) \end{array} \]
- Answer

Problem 4

You are given two ordered bases \(\EuScript{B}=(\Vect{b}_1,\dots ,\Vect{b}_n)\) and \(\EuScript{C}=(\Vect{c}_1,\dots ,\Vect{c}_n)\) of \(\RNrSpc{n}\). For each of the following statements decide if it is true or false

The matrix which converts from \(\EuScript{C}\)-coordinates to \(\EuScript{B}\)-coordinates is denoted \(\Mtrx{C}_{\EuScript{B}\EuScript{C}}\).
- Answer
The matrix which converts from \(\EuScript{C}\)-coordinates to \(\EuScript{B}\)-coordinates has size \((3,3)\).
- Hint Hint Answer
The matrix which converts from \(\EuScript{B}\)-coordinates to \(\EuScript{C}\)-coordinates is invertible.
- Solution
The matrix which converts from \(\EuScript{B}\)-coordinates to \(\EuScript{C}\)-coordinates has rank \(n\).
- Hint Solution
The determinant of the matrix which converts from \(\EuScript{B}\)-coordinates to \(\EuScript{C}\)-coordinates is nonzero.
- Hint Solution

Solution for Problem 2 a

We construct the required coordinate conversion matrix conveniently as

\(\Mtrx{C}_{\EuScript{C}\EuScript{B}}\)

\(=\)

\(\Mtrx{C}_{\EuScript{C}\EuScript{S}} \Mtrx{C}_{\EuScript{S}\EuScript{B}}\)

This is indeed convenient because we know

\[ \Mtrx{C}_{\EuScript{S}\EuScript{B}} = \left[ \begin{array}{rrr} 1 & -1 & 0 \\ 0 & -1 & -1 \\ 1 & 0 & -1 \end{array} \right] \]

and

\[ \Mtrx{C}_{\EuScript{C}\EuScript{S}} = \left( \Mtrx{C}_{\EuScript{S}\EuScript{C}}\right)^{-1} = \left[ \begin{array}{rrr} 1 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 1 & 2 \end{array} \right]^{-1} \]

With the row reduction method, we obtain

\[ \left[ \begin{array}{rrr} 1 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 1 & 2 \end{array} \right]^{-1} = \dfrac{1}{2} \left[ \begin{array}{rrr} 2 & -2 & 0 \\ 0 & -2 & 0 \\ 0 & 1 & 1 \end{array} \right] \]

Therefore we obtain

\[ \Mtrx{C}_{\EuScript{C}\EuScript{B}} = \Mtrx{C}_{\EuScript{C}\EuScript{S}} \cdot \Mtrx{C}_{\EuScript{S}\EuScript{B}} = \dfrac{1}{2}\left[ \begin{array}{rrr} 2 & 0 & 2 \\ 0 & 2 & 2 \\ 1 & -1 & -2 \end{array} \right] \]