16 | February | 2012

always it amuses me the incredulity of people when they receive an explanation of the covariant and contravariant behavior of linear algebra matters.

If one has a change of basis $\mathbb{R}^2\to\mathbb{R}^2$ , you gotta to specify the basis that is assigned.

If you begin by chosing the canonical basis $e_1,e_2$ , where

$e_1=\left(\!\!\begin{array}{c}1\\ 0\end{array}\!\!\right)$ , $e_2=\left(\!\!\begin{array}{c}0\\ 1\end{array}\!\!\right)$ ,

and another basis, say

$b_1=\left(\!\!\begin{array}{c}1\\ 0\end{array}\!\!\right)$ , $b_2=\left(\!\!\begin{array}{c}1\\ 1\end{array}\!\!\right)$ ,

then we have $b_1=e_1$ and $b_2=e_1+e_2$ . From here one can resolve: $e_1=b_1$ and $e_2=-b_1+b_2$ .

So if a vector $v=Ae_1+Be_2$ is an arbitrary (think… $A,B\in\mathbb{R}$ ) element in $\mathbb{R}^2$ then its expression in the new basis is:

$v=Ab_1+B(-b_1+b_2)$

$=(A-B)b_1+Bb_2$ .

Remember $A,B\in\mathbb{R}$ .

So, if $v=\left(\!\!\begin{array}{c}A\\ B\end{array}\!\!\right)_e$ are the component in the old basis, then $v=\left(\!\!\begin{array}{c}A-B\\ B\end{array}\!\!\right)_b$ are the components in the new one.

Matricially what we see is this:

$\left(\!\!\begin{array}{cc}1&-1\\ 0&1\end{array}\!\!\right)\left(\!\!\begin{array}{c}A\\ B\end{array}\!\!\right)=\left(\!\!\begin{array}{c}A-B\\ B\end{array}\!\!\right)$ .