Tag Archives: change of basis

solution to a visual quizz


the correction is:

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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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