Changing a vector's basis can be done using *vector projections* or *matrix transformations*. They are equivalent when the basis vectors are [[Linearly Independent Vectors]]. ## Changing Basis in Vector Form Let the vector $r_e$ in the standard basis vectors $(e_i, e_j)$ be: $ r_e = \begin{bmatrix} 3 \\ 4 \end{bmatrix}, \quad e_1= \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad e_2= \begin{bmatrix} 0 \\ 1 \end{bmatrix} $ Suppose we want to express $r_e$ in terms of new basis vectors $(b_1, b_2)$: $ b_1= \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad b_2= \begin{bmatrix} -2 \\ \phantom-4 \end{bmatrix} $ ![[basis-vectors-1.png|center|350]] *Step 1: Check Orthogonality of New Basis Vectors* To apply vector projection for this re-indexation, the new basis vectors $b_1, b_2$ must be orthogonal. This is the case when the [[Dot Product]] equals zero. $ b_1 \cdot b_2= \begin{bmatrix} 2 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} -2 \\ \phantom-4 \end{bmatrix}=2*(-2)+1*4=0 $ *Step 2: Project Vector onto New Basis* We use scalar projection to find the length of the components of $r_e$ along each dimension $b_1$ and $b_2$. $ \lvert \text{adj}_1 \rvert =\frac{b_1 \cdot r_e}{\lvert b_1 \rvert}, \quad \lvert \text{adj}_2 \rvert =\frac{b_2 \cdot r_e}{\lvert b_2 \rvert} $ Projection onto $b_1$: $ \vert \text{adj}_1 \vert = \frac{\begin{bmatrix}2 \\1\end{bmatrix} \cdot \begin{bmatrix}3 \\ 4\end{bmatrix}}{2^2+1^2} = \frac{10}{\sqrt 5}$ Projection onto $b_2$: $ \vert \text{adj}_2 \vert = \frac{\begin{bmatrix}-2 \\\phantom{..}4\end{bmatrix} \cdot \begin{bmatrix}3 \\4 \end{bmatrix}}{-2^2+4^2} = \frac{10}{\sqrt {20}} $ *Step 3: Express Vector in the new Basis* Dividing the [[Vector Length]] of e.g. $\text{adj}_1$ by the length of the new basis vectors $b_1$, tells us what fraction of $b_1$ we have to go along $b_1$. $ r_b = \begin{bmatrix} \frac{10}{\sqrt 5} * \frac{1}{\sqrt 5}\\[6pt] \frac{10}{\sqrt {20}} * \frac{1}{\sqrt {20}} \end{bmatrix} = \begin{bmatrix} 2\\ 0.5 \end{bmatrix} $ We get that the vector $r$ expressed in basis vectors $(b_1, b_2)$ point: - $2$ units of $b_1$ into direction of $b_1$ - $0.5$ units of $b_2$ into direction of $b_2$ ## Changing Basis in Matrix Form When we introduce a new set of basis vectors $b_1, b_2$, we can represent them as columns of a matrix $\mathbf B$. $ b_1 = \begin{bmatrix} 2 \\ 10\end{bmatrix}, \quad b_2 = \begin{bmatrix} 3 \\ 1\end{bmatrix}, \quad \mathbf B= \begin{bmatrix} 2 & 3 \\ 10 & 1 \end{bmatrix} $ **From New Basis to Standard Basis:** To re-express a vector $r_b$ (with basis $b_1, b_2$) in terms of standard [[Basis Vectors]], we multiply the matrix $\mathbf B$ with $r_b$. $ r_e= \mathbf Br_b$ For example: $ r_b=\begin{bmatrix}3\\2\end{bmatrix}, \quad r_e=\mathbf B r_b = \begin{bmatrix} 2 & 3 \\ 10 & 1 \end{bmatrix}* \begin{bmatrix} 3 \\2 \end{bmatrix} = \begin{bmatrix} 12 \\32 \end{bmatrix} $ **From Standard Basis to New Basis:** When we have a vector $r_e$ (with standard Basis vectors), and need to re-express it in terms of the new basis, we multiply it with the [[Inverse Matrix]] of $\mathbf B$. $r_b= \mathbf B^{-1}r_e$ For example: $ r_e=\begin{bmatrix}3\\2\end{bmatrix}, \quad r_b=\mathbf B^{-1} r_e = \frac{1}{\det (\mathbf B)}*\begin{bmatrix} 1 & -3 \\ -10 & 2 \end{bmatrix}* \begin{bmatrix} 3 \\2 \end{bmatrix}$ >[!note:] >In contrast to vector projection, this approach also works when the new basis vectors are not orthogonal.