Tag Archives: linear algebra

Einstein-Penrose ‘s strong sum convention

the rank one tensors’ basis changes



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wedge product example

When bivectors are defined by


so, for two generic covectors

\theta=a\beta^1+b\beta^2+c\beta^3 and \phi=d\beta^1+e\beta^2+f\beta^3,

we have the bivector




Cf. this with the data \left(\begin{array}{c}a\\b\\c\end{array}\right) and \left(\begin{array}{c}d\\e\\f\end{array}\right) to construct the famous


So, nobody should be confused about the uses of the symbol \wedge dans le calcul vectoriel XD


Filed under algebra, cucei math, differential geometry, math analysis, mathematics, multilinear algebra, what is math, word algebra

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:



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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simu exam comment…

The inversion of a matrix.

For a change of basis in \mathbb{R}^3:




we have as a change-of-basis-matrix:

\left(\begin{array}{ccc}1&1&0\\ 0&2&1\\ 0&0&-1\end{array}\right)

and by solving for:




then we get as an inverse of that matrix:

\left(\begin{array}{ccc}1&-1/2&-1/2\\ 0&1/2&1/2\\ 0&0&-1\end{array}\right)

All that inside a recent exam did at the dept of maths

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real quadratic forms

A quadratic form in an m -dimensional  real vectorspace V, is a bilineal map V\times V\to\mathbb{R} which can be determined by a m\times m-matrix A via

(v,w)\mapsto v^TAw

If we are allowed to write g(v,w)=v^TAw we can find that he (or she) possibly satisfy

  1. g(v,w)=g(w,v),  property dubbed symmetry
  2. g(v,v)\ge0, positive-definiteness
  3. g(v,v)=0 iff v=0, non-degeneracy

In case of affirmatively both three are satisfy g is called a metric on V and the pair


is called (real) Euclidean vectorspace…

MORE on \mathbb{R}^2

MORE on \mathbb{R}^3


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a GPS for a surface to do calculus on it


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matrices especiales de 2×2, módulo 2 y módulo 3

no es difícil calcular que SL_2(\mathbb{Z}_2) tiene seis elementos y que este grupo es el grupo simétrico S_3=\langle a,b\mid a^2=b^3=e,\ ab^2=ba\rangle.

¿Y que hay acerca de SL_2(\mathbb{Z}_3)? también no es difícil calcular |SL_2(\mathbb{Z}_3)|=24.

¿Es cierto qué SL_2(\mathbb{Z}_3)=S_6?

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