# change of basis and change of components

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:

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

This corresponds to the relation

$M^{-1}v_e=v_b, \qquad (*)$

which -contrasted against $Me_i=b_i$ in the change between those basis- causes some mind twist  :D…  :p

One says then that the base vectors co-vary, but components of vectors contra-vary.

For more: Covariance and contravariance of vectors.

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### One response to “change of basis and change of components”

1. kt

Matriz de una transformación lineal y la de su “pullback”

Si $[A]=[{A^i}_j]$ es la matriz de una transformación lineal
$A:V_b\to W_c$ entonces
¿cuál es la matriz de su pullback $A^*:W_{\gamma}^*\to V_{\beta}^*\quad?$

Afirmamos que

$[A^*]=[{({A^{\top}})^i}_j]$

donde
$A^{\top}$ es la matriz transpuesta de $A$.

Demostración

$A:V_b\to W_c$

$b_i\longmapsto {A^s}_ic_s$

$V_b\stackrel{A}\to W_c\stackrel{f}\to {\Bbb{R}}$

$V_b\stackrel{f\circ A}\longrightarrow {\Bbb{R}}$

$A^*:W_{\gamma}^*\to V_{\beta}^*$

$f\longmapsto A^*f=f\circ A$

$\gamma^k\longmapsto A^*\gamma^k=\gamma^k\circ A$

$\gamma^k\longmapsto A^*\gamma^k={B_s}^k\beta^s$

$A^*\gamma^k(b_i)={B_s}^k\beta^s(b_i)={B_s}^k{\delta^s}_i={B_i}^k$

$A^*\gamma^k(b_i)=\gamma^k\circ A(b_i)=\gamma^k({A^s}_ic_s)$

$A^*\gamma^k(b_i)={A^s}_i\gamma^k(c_s)={A^s}_i{\delta^k}_s={A^k}_i$

Por lo tanto ${B_i}^k={A^k}_i$

y así

$[A^*]=[A]^{\top}$

$\Box$