Change of Basis - Examples with Solutions

Change of Coordinates Matrices

Given two bases for a vector space V , the change of coordinates matrix from the basis B to the basis A is defined as [1] where are the column vectors expressing the coordinates of the vectors with respect to the basis A .
In a similar way is defined by It can be shown that

Applications of Change of Coordinates Matrices

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