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
are defined above
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 \( P_{B \leftarrow A} \cdot [x]_A = [x]_B\)
