26. Change of Basis
Dated: 11-06-2025
Example
Consider 2 basis \(\mathbb B = \{\vec b_1, \vec b_2\}\) and \(\mathbb C = \{\vec c_1, \vec c_2\}\) for a vector space1 \(\mathbb V\) such that \(\vec b_1 = 4 \vec c_1 + \vec c_2\) and \(\vec b_2 = -6 \vec c_1 + \vec c_2\).
Suppose that
\[\vec x = 3 \vec b_1 + \vec b_2\]
\[
\implies [\vec x]_{\mathbb B} =
\begin{bmatrix}
3 \\
1
\end{bmatrix}
\]
Find \([\vec x]_{\mathbb C}\).
Solution
\[[\vec x]_{\mathbb C} = [3 \vec b_1 + \vec b_2]_{\mathbb C}\]
\[= 3[\vec b_1]_{\mathbb C} + [\vec b_2]_{\mathbb C}\]
\[
=
\begin{bmatrix}
[\vec b_1]_{\mathbb C} & [\vec b_2]_{\mathbb C}
\end{bmatrix}
\begin{bmatrix}
3 \\
1
\end{bmatrix}
\]
\[
\because
[\vec b_1]_{\mathbb C} =
\begin{bmatrix}
4\\
1
\end{bmatrix}
\quad
[\vec b_2]_{\mathbb C} =
\begin{bmatrix}
-6\\
1
\end{bmatrix}
\]
\[
[\vec x]_{\mathbb C} =
\begin{bmatrix}
4 & -6\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
3\\
1
\end{bmatrix}
=
\begin{bmatrix}
6\\
4
\end{bmatrix}
\]
Theorem
Let \(\mathbb B = \{\vec b_1, \ldots, \vec b_n\}\) and \(\mathbb C = \{\vec c_1, \ldots, \vec c_n\}\) be bases of a vector space1 \(\mathbb V\) then there exists \(n \times n\) matrix2 \(P_{\mathbb C \leftarrow \mathbb B}\) such that
\[[\vec x]_{\mathbb C} = P_{\mathbb C \leftarrow \mathbb B}[\vec x]_\mathbb B\]
Where \(P_{\mathbb C \leftarrow \mathbb B}\) is called change-of-coordinates matrix2 from \(\mathbb B\) to \(\mathbb C\).
\[
P_{\mathbb C \leftarrow \mathbb B} =
\begin{bmatrix}
[\vec b_1]_\mathbb C & [\vec b_2]_\mathbb C & \cdots & [\vec b_n]_\mathbb C
\end{bmatrix}
\]
Commutative diagram for basis transformations.
\[\left(P_{\mathbb C \leftarrow \mathbb B}\right)^{-1} = P_{\mathbb B \leftarrow \mathbb C}\]
References
Read more about notations and symbols.