# Vectors and matrices: performing a change of basis

We set the following notation for change of basis. Consider basis $\{\mb{b}_1, \mb{b}_2\}$, and we express $\mb{x} = 3 \mb{b}_1 + \mb{b}_2$. Then we express $\mb{x}$ using the B-coordinate vector as $[\mb{x}]_B = \begin{bmatrix} 3\\ 1\end{bmatrix}$

### Example

We give an example of how to change basis. Consider two bases $B = \{\mb{b}_1, \mb{b}_2\}$ and $C = \{\mb{c}_1, \mb{c}_2\}$ where $\mb{b}_1 = 4\mb{c}_1 + \mb{c}_2, \qquad \mb{b}_2 = -6 \mb{c}_1 + \mb{c}_2.$

Suppose that $x = 3 \mb{b}_1 + \mb{b}_2$. What is $x$ under the basis C? We find $[\mb{x}]_C = [3 \mb{b}_1 + \mb{b}_2]_C = 3[\mb{b}_1]_C + [\mb{b}_2]_C = 3\begin{bmatrix}4 \\ 1\end{bmatrix} + \begin{bmatrix}-6 \\ 1 \end{bmatrix}.$

Note that we could have equally written this as $[\mb{x}]_C = \begin{bmatrix} 4 & -6 \\ 1 & 1\end{bmatrix} \begin{bmatrix} 3 \\ 1\end{bmatrix} = \begin{bmatrix} 6 \\ 4\end{bmatrix}$

### Change of basis theorem

Let $B = \{\mb{b}_1, \mb{b}_2, \ldots, \mb{b}_n\}$ and $C = \{\mb{c}_1, \mb{c}_2, \ldots \mb{c}_n\}$ be bases. The change of basis formula follows from \begin{equation} [\mb{x}]_C = P [\mb{x}]_B, \end{equation}

where the columns of $P$ are the $C$-coordinate vectors of the basis $\mb{b}_i$, that is $P = \begin{bmatrix} [\mb{b}_1]_C & [\mb{b}_2]_C & \cdots & [\mb{b}_n]_C \end{bmatrix}$ 