# Change of Basis - Examples with Solutions

   

## Change of Coordinates Matrices

Given two bases $A = \{\textbf{a}_1 , \textbf{a}_2 , ... , \textbf{a}_n \}$ and $B = \{\textbf{b}_1 , \textbf{b}_2 , ... , \textbf{b}_n \}$ for a vector space $V$, the change of coordinates matrix from the basis $B$ to the basis $A$ is defined as [1] $P_{A \leftarrow B} = \left[ \;\;[ \textbf{b}_1]_A \;\; [\textbf{b}_2]_A ... [\textbf{b}_n]_A \;\; \right]$ where $[\textbf{b}_1]_A$ , $[\textbf{b}_1]_A$ ... $[\textbf{b}_n]_A$ are the column vectors expressing the coordinates of the vectors $\textbf{b}_1$ , $\textbf{b}_2$ ... $\textbf{b}_2$ with respect to the basis $A$.
In a similar way $P_{B \leftarrow A}$ is defined by $P_{B \leftarrow A} = \left[ \;\;[ \textbf{a}_1]_B \;\; [\textbf{a}_2]_B ... [\textbf{a}_n]_B \;\; \right]$ It can be shown that $P_{B \leftarrow A} = (P_{A \leftarrow B})^{-1}$

## Applications of Change of Coordinates Matrices

If $A$ and $B$ are bases for a vector space $V$ and $\textbf{x}$ is a vector in $V$, then [1]
$[x]_A = P_{A \leftarrow B} [x]_B$ and $[x]_B = P_{B \leftarrow A} [x]_A$
where $P_{A \leftarrow B}$ and $P_{B \leftarrow A}$ are defined above.

## Examples with Solutions

Example 1
Given the bases $A = \left\{ \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} , \begin{bmatrix} -2 \\ -3 \\ \end{bmatrix} \right\}$ and $B = \left\{ \begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} , \begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} \right\}$ for a vector space $V$,
a) find matrix $P_{A \leftarrow B}$
b) find matrix $P_{B \leftarrow A}$
c) show that matrices $P_{A \leftarrow B}$ and $P_{B \leftarrow A}$ are inverse of each other.

Solution to Example 1
Let $A = \left\{ \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} , \begin{bmatrix} -2 \\ -3 \\ \end{bmatrix} \right\} = \{ a_1 , a_2 \}$ and $B = \left\{ \begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} , \begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} \right\} = \{ b_1 , b_2 \}$
a)
$P_{A \leftarrow B} = \left[ \;\;[ \textbf{b}_1]_A \; \; [\textbf{b}_2]_A \; \; \right]$
Let $k_1$, $k_2$, $k'_1$, $k'_2$ be constants such that
$b_1 = k_1 a_1 + k_2 a_2$ and $b_2 = k'_1 a_1 + k'_2 a_2$       (I)
Hence
$[ \textbf{b}_1]_A = \begin{bmatrix} k_1 \\ k_2 \\ \end{bmatrix}$ and $[ \textbf{b}_2]_A = \begin{bmatrix} k'_1 \\ k'_2 \\ \end{bmatrix}$
We now need to find $k_1$, $k_2$, $k'_1$, $k'_2$ by solving the equations in (I) above which in matrix form written as
$[ a_1 \quad a_2 ] \begin{bmatrix} k_1 & k'_1\\ k_2 & k_2\\ \end{bmatrix} = [b_1 \quad b_2]$
The augmented matrix of the above is
$\begin{bmatrix} 1 & -2 \; | \; 2 & 1\\ 2 & -3 \; | \; 1 & 3\\ \end{bmatrix}$
Row reduce the above to
$\begin{bmatrix} 1 & 0 \; | \; -4 & 3\\ 0 & 1 \; | \; -3 & 1\\ \end{bmatrix}$           (II)
Hence
$P_{A \leftarrow B} = \begin{bmatrix} k_1 & k'_1\\ k_2 & k_2\\ \end{bmatrix} = \begin{bmatrix} -4 & 3\\ -3 & 1\\ \end{bmatrix}$
Note that omitting the details above, to find the change of coordinates matrix $P_{A \leftarrow B}$, row reduce the matrix $[ A \; | \; B ]$, formed by the columns of basis $A$ and the columns of basis $B$, to the form $[ I \; | \; P_{A \leftarrow B} ]$ where $I$ is the identity matrix as shown above in (II).

b)
In a similar way as above, but omitting the details, we find the change of coordinates matrix $P_{B \leftarrow A}$ in two steps:
1)     form the matrix $[ B \; | \; A ]$ using columns of basis $B$ and the columns of basis $A$ as follows
$\begin{bmatrix} 2 & 1 \; | 1 & -2 \\ 1 & 3 \; | \; 2 & -3 \\ \end{bmatrix}$
2)     row reduce the above to obtain
$\begin{bmatrix} 1 & 0 \; | \; \dfrac{1}{5} & -\dfrac{3}{5}\\ 0 & 1 \; | \; \dfrac{3}{5} & -\dfrac{4}{5}\\ \end{bmatrix}$
Having row reduced matrix $[ B \; | \; A ]$ to the form $[ I \; | \; P_{B \leftarrow A} ]$, we can write
$P_{B \leftarrow A} = \begin{bmatrix} \dfrac{1}{5} & -\dfrac{3}{5}\\ \dfrac{3}{5} & -\dfrac{4}{5}\\ \end{bmatrix}$

c)
To show that $P_{A \leftarrow A}$ and $P_{B \leftarrow B}$ are inverse of each oether, we need to show that their products are equal to the identity matrix.
$P_{A \leftarrow A} \times P_{B \leftarrow A} = \begin{bmatrix} -4 & 3\\ -3 & 1\\ \end{bmatrix} \times \begin{bmatrix} \dfrac{1}{5} & -\dfrac{3}{5}\\ \dfrac{3}{5} & -\dfrac{4}{5}\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}$
and
$P_{B \leftarrow A} \times P_{A \leftarrow A} = \begin{bmatrix} \dfrac{1}{5} & -\dfrac{3}{5}\\ \dfrac{3}{5} & -\dfrac{4}{5}\\ \end{bmatrix} \times \begin{bmatrix} -4 & 3\\ -3 & 1\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}$

Example 2
Given the bases $A = \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 0\\ 2\\ -1 \end{bmatrix} , \begin{bmatrix} 1 \\ -1\\ -1\\ 2 \end{bmatrix} \right \}$ and $B = \left\{ \begin{bmatrix} 2 \\ 0 \\ -1 \\ 2 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 3 \\ -3 \end{bmatrix} , \begin{bmatrix} 3 \\ 1 \\ 4 \\ -2 \end{bmatrix} \right\}$ for a vector space $V$ and vector $x = \begin{bmatrix} 5 \\ 2 \\ 6 \\ -3 \end{bmatrix}$ in $V$.
a) Find matrix $P_{A \leftarrow B}$
b) Find matrix $P_{B \leftarrow A}$
c) Show that matrices $P_{A \leftarrow B}$ and $P_{B \leftarrow A}$ are inverse of each other.
d) Find $[x]_A$ and $[x]_B$ directly using the definition of coordinates.
e) Verify the formulas $[x]_A = P_{A \leftarrow B} [x]_B$ and $[x]_B = P_{B \leftarrow A} [x]_A$ given above.

Solution to Example 2
a)
$P_{A \leftarrow B}$ is found by row reducing the augmented matrix (see example 1 above)
$\begin{bmatrix} 1 & 1 & 1 &|& 2 & 0 & 3\\ 1 & 0 & -1 &|& 0 & 1 & 1 \\ 0 & 2 & -1 &|& -1 & 3 & 4\\ 0 & -1 & 2 &|& 2 & -3 & -2 \end{bmatrix}$
Row reduce to
$\begin{bmatrix} 1 & 0 & 0 &|& 1 & 0 & 1\\ 0 & 1 & 0 &|& 0 & 1 & 2 \\ 0 & 0 & 1 &|& 1 & -1 & 0\\ 0 & 0 & 0 &|& 0 & 0 & 0 \end{bmatrix}$
Hence
$P_{A \leftarrow B} = \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2 \\ 1 & -1 & 0\\ \end{bmatrix}$

b)
$P_{B \leftarrow A}$ is found by row reducing the augmented matrix (see example 1 above)
$\begin{bmatrix} 2 & 0 & 3 &|& 1 & 1 & 1\\ 0 & 1 & 1 &|& 1 & 0 & -1 \\ -1 & 3 & 4 &|& 0 & 2 & -1 \\ 2 & -3 & -2 &|& 0 & -1 & 2 \end{bmatrix}$
Row reduce to
$\begin{bmatrix} 1 & 0 & 0 &|& 2 & -1 & -1\\ 0 & 1 & 0 &|& 2 & -1 & -2 \\ 0 & 0 & 1 &|& -1 & 1 & 1\\ 0 & 0 & 0 &|& 0 & 0 & 0 \end{bmatrix}$
Hence
$P_{B \leftarrow A} = \begin{bmatrix} 2 & -1 & -1\\ 2 & -1 & -2 \\ -1 & 1 & 1 \end{bmatrix}$

c)
It is enough to show that their product of $P_{A \leftarrow B}$ and $P_{B \leftarrow A}$ gives the identity matrix.
$\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2 \\ 1 & -1 & 0\\ \end{bmatrix} \times \begin{bmatrix} 2 & -1 & -1\\ 2 & -1 & -2 \\ -1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -1\\ 2 & -1 & -2 \\ -1 & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2 \\ 1 & -1 & 0\\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{bmatrix}$

d)
To find the coordinates of $x$ in base $A$ written as $[x]_A$, we need to row reduce
$\begin{bmatrix} 1 & 1 & 1 &|& 5\\ 1 & 0 & -1 &|& 2 \\ 0 & 2 & -1 &|& 6 \\ 0 & -1 & 2 &|& - 3 \end{bmatrix}$
Row reduce to solve
$\begin{bmatrix} 1 & 0 & 0 &|& 2\\ 0 & 1 & 0 &|& 3 \\ 0 & 0 & 1 &|& 0 \\ 0 & 0 & 0 &|& 0 \end{bmatrix}$
Hence
$[x]_A = \begin{bmatrix} 2\\ 3 \\ 0 \\ \end{bmatrix}$

To find the coordinates of $x$ in base $B$ written as $[x]_B$, we need to row reduce
$\begin{bmatrix} 2 & 0 & 3 &|& 5\\ 0 & 1 & 1 &|& 2 \\ -1 & 3 & 4 &|& 6 \\ 2 & -3 & -2&|& - 3 \end{bmatrix}$
Row reduce to solve
$\begin{bmatrix} 1 & 0 & 0 &|& 1\\ 0 & 1 & 0 &|& 1 \\ 0 & 0 & 1 &|& 1 \\ 0 & 0 & 0 &|& 0 \end{bmatrix}$
Hence
$[x]_B = \begin{bmatrix} 1\\ 1 \\ 1 \\ \end{bmatrix}$

e)
We compute the product $P_{A \leftarrow B} \cdot [x]_B$
$\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2 \\ 1 & -1 & 0\\ \end{bmatrix} \cdot \begin{bmatrix} 1\\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 2\\ 3 \\ 0 \\ \end{bmatrix}$
and we can see that $P_{A \leftarrow B} \cdot [x]_B = [x]_A$

We compute the product $P_{B \leftarrow A} \cdot [x]_A$
$\begin{bmatrix} 2 & -1 & -1\\ 2 & -1 & -2 \\ -1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2\\ 3 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 1\\ 1 \\ 1\\ \end{bmatrix}$
and we can see that $P_{B \leftarrow A} \cdot [x]_A = [x]_B$

## More References and links

1. Linear Algebra - Questions with Solutions
2. Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
3. Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
4. Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
5. Linear Algebra with Applications - 2012 - Gareth Williams
6. Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres