> [!Note] Definition (Change of basis matrix) > Suppose $e_{1},\dots ,e_{n}$ is the standard basis and $\underline{f}_{1},\dots,\underline{f}_{n}$ is any basis of $\mathbb{R}^{n}$. > > Then for all $\underline{w}\in\mathbb{R}^{n}$, there are unique scalars $\lambda_{i}$ and $\mu_{i}$ for which $\begin{array}{rcl}\underline{w}&=&\lambda_1\underline{e}_1+\ldots+\lambda_n\underline{e}_n&\text{and}\\\underline{w}&=&\mu_1\underline{f}_1+\ldots+\mu_n\underline{f}_n\end{array}$ > >Using matrices we can rewrite these as $Q(\mu_{1},\dots ,\mu_{n})^{T}= (\lambda_{1},\dots,\lambda_{n})^{T}$ where the columns of $Q$ are $\underline{f}_{1},\dots,\underline{f}_{n}$. > >Then $(\mu_{1},\dots \mu_{n})=Q^{-1} (\lambda_{1},\dots,\lambda_{n})^{T}$. > >$Q$ is known as the change of basis matrix. > [!NOTE] Theorem (Change of basis formula for linear maps) > Suppose $\varphi:V\to W$ is a [[Linear Map|linear map]] between [[Finite Dimensional Vector Space|FDVS]]. Suppose we have: > 1. bases $\mathcal{A}=\{ v_{1},\dots,v_{n} \}$ of $V$ and $\mathcal{B}=\{ w_{1},\dots, w_{m} \}$ of $W,$ so that with respect to these bases $\varphi$ is the [[Left Multiplication Linear Map of Real Matrix#^2b50ef|left multiplication linear map wrt to the given bases]] of some matrix $A\in \mathbb{R}^{mn}.$ > 2. an alternative basis $\mathcal{A'}=\{ v_{1}', \dots,v_{n}' \}$ of $V$ with change of basis matrix $Q$ for which $\chi_{\mathcal{A}}=L_{Q} \circ \chi_{\mathcal{A}'}$ > 3. an alternative basis $\mathcal{B'}=\{ w_{1},\dots,w_{m}' \}$ of $W$ with change of basis matrix $P$ for which $\chi_{\mathcal{B}}=L_{P} \circ \chi_{\mathcal{B}}$ > >Then with respect to bases $\mathcal{A}'$ of $V$ and $\mathcal{B}'$ of $W,$ the maps $\varphi$ is represent by the matrix $B=P^{-1}A Q$ which is [[Equivalence of Matrices|equivalent]] to $A.$ Change of basis matrix is given by $Q=\begin{pmatrix} \frac{1}{\sqrt{ 2 }} & \frac{1}{\sqrt{ 2 }} & 0 \\ \frac{1}{\sqrt{ 2 }} & -\frac{1}{\sqrt{ 2 }} &0 \\ 0 & 0 & 1 \end{pmatrix}$ Row reduce $(Q\mid I_{3})$ $\begin{align} &\left(\begin{array}{ccc|ccc} \frac{1}{\sqrt{ 2 }} & \frac{1}{\sqrt{ 2 }} & 0 & 1 & 0 & 0 \\ \frac{1}{\sqrt{ 2 }} & -\frac{1}{\sqrt{ 2 }} &0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right) \stackrel{M_{1}(\sqrt{ 2 }, M_{2}(\sqrt{ 2 }))}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 1 & 0 & \sqrt{ 2 } & 0 & 0 \\ 1 & -1 &0 & 0 & \sqrt{ 2 } & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right) \\ &\stackrel{A_{12}(-1),M_{2}\left( -\frac{1}{2} \right)}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 1 & 0 & \sqrt{ 2 } & 0 & 0 \\ 0 & 1 &0 & \frac{\sqrt{ 2 }}{2} & -\frac{\sqrt{ 2 }}{2} & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right) \stackrel{A_{21} (-1)}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{\sqrt{ 2 }}{2} & \frac{\sqrt{ 2 }}{2} & 0 \\ 0 & 1 &0 & \frac{\sqrt{ 2 }}{2} & -\frac{\sqrt{ 2 }}{2} & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right). \end{align}$So $\begin{pmatrix} \lambda_{1} \\ \lambda_{2} \\ \lambda_{3} \end{pmatrix} = Q^{-1} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \frac{\sqrt{ 2 }}{2} \begin{pmatrix} 1 & 1 & 0 \\ 1 & - 1 & 0 \\ 0 & 0 & \sqrt{ 2 } \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \frac{\sqrt{ 2 }}{2} \begin{pmatrix} 0 \\ 0 \\ \sqrt{ 2 } \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$