Tag Archives: linear algebra

Einstein-Penrose ‘s strong sum convention

the rank one tensors’ basis changes Filed under math, mathematics, multilinear algebra, word algebra

wedge product example

When bivectors are defined by $\beta^i\wedge\beta^j:=\beta^i\otimes\beta^j-\beta^j\otimes\beta^i$,

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 $\theta\wedge\phi=(bf-ce)\beta^2\wedge\beta^3+(cd-af)\beta^3\wedge\beta^1+(ad-be)\beta^1\wedge\beta^2$.

Otherwise,

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 $\left(\begin{array}{c}a\\b\\c\end{array}\right)\times\left(\begin{array}{c}d\\e\\f\end{array}\right)=\left(\begin{array}{c}bf-ce\\cd-af\\ad-be\end{array}\right)$

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

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.

1 Comment

Filed under algebra, multilinear algebra

simu exam comment…

The inversion of a matrix.

For a change of basis in $\mathbb{R}^3$: $\eta_1=\varepsilon_1$ $\eta_2=\varepsilon_1+2\varepsilon_2$ $\eta_3=\varepsilon_2-\varepsilon_3$

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: $\varepsilon_1=\eta_1$ $\varepsilon_2=-\frac{1}{2}\eta_1+\frac{1}{2}\eta_2$ $\varepsilon_2=-\frac{1}{2}\eta_1+\frac{1}{2}\eta_2-\eta_3$

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

Filed under algebra, mathematics, multilinear algebra, what is mathematics

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 $(V,g)$

is called (real) Euclidean vectorspace…

MORE on $\mathbb{R}^2$

MORE on $\mathbb{R}^3$

Filed under math, multilinear algebra

a GPS for a surface to do calculus on it 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 $SL_2(\mathbb{Z}_3)=S_6$?