Change of Basis

April 17, 2019

Chang of basis for a vector $\mathbf p$ representing coefficients/coordinates for the old basis into a new vector of coordinates $\mathbf c$ with base $W$ is

$\mathbf p=W\mathbf c\\ \mathbf c=W^{-1}\mathbf p$

Good Basis is

Fast at getting inverses
Fast at multiplying a vector
Good for compression, some columns do not matter that much

Change of Basis w/ Linear Transformation

Let the transformation $T(·)$ has matrix $A$ in $\mathbb R^n$ with basis $v_1,…v_n$ , and it transforms into $\mathbb R^m$ with basis $w_1,…,w_m$ . So $A:m\times n$ . We know $T(·)$ completely if know its output on every base vectors: $T(v_1),T(v_2),…,T(v_n)$ . Let’s say we have a vector $\mathbf x=c_1v_1+c_2v_2+…+c_nv_n$ , then $T(\mathbf x)=c_1T(v_1)+…+c_nT(v_n)$ . Along with $A=T(·)$ , we need to make sure $A\mathbf x=c_1Av_1+c_2Av_2+…+c_nAv_n$ can produce the same effect of $T(\mathbf x)$ . Let

$A=\begin{bmatrix} a_{11} & ... & a_{1n}\\ \vdots&\times&\vdots\\ a_{m1}&...&a_{mn}\\ \end{bmatrix}$

and $T(v_1)=T(\begin{bmatrix} 1\\0\\\vdots \end{bmatrix})$ because it's only the first vector in the basis $\begin{bmatrix} \vert & ... & \vert\\ v_1&…&v_n\\ \vert&…&\vert\\ \end{bmatrix}$ . Therefore $Av_1$ is equal to the first column of $A$ multiplying $v_1$ . Thus $a_{11}…a_{m1}$ satisfy $a_{11}w_1+…a_{m1}w_m=T(v_1)$ (just the $a$ the coefficients matters, $w_i$ are unknowns).