Is (u,v,u^v) a Positive Basis in Vector Algebra?

[LCSphysicist]

Homework Statement

If (u,v,w) is a positive basis, so (u^v, v^w,w^u) is too.

Relevant Equations

All below

I think we can say that (u,v,u^v) is a positive basis, so as (w^v,v,w) and (u,w^u,w). (1)

So
u^v = βw
v^w = γu
w^u = λv

where λ, β, and γ > 0 (*)

(u^v, v^w,w^u) = (βw,γu,λv)

\begin{vmatrix}
0 & 0 & β \\
γ & 0 & 0 \\
0 & λ & 0 \\
\end{vmatrix}

This determinant is positive by (*)

What you think about?

 

[Mark44]

Homework Statement:: If (u,v,w) is a positive basis, so (u^v, v^w,w^u) is too.
Relevant Equations:: All below

I think we can say that (u,v,u^v) is a positive basis, so as (w^v,v,w) and (u,w^u,w). (1)

What does it mean for a basis to be positive?
Also, assuming u, v, and w are vectors, what do the expressions u^v, v^w, and w^u mean? I don't know what it means to raise a vector to a vector power.

So
u^v = βw
v^w = γu
w^u = λv

where λ, β, and γ > 0 (*)

(u^v, v^w,w^u) = (βw,γu,λv)

\begin{vmatrix}
0 & 0 & β \\
γ & 0 & 0 \\
0 & λ & 0 \\
\end{vmatrix}

This determinant is positive by (*)

What you think about?

 

[LCSphysicist]

What does it mean for a basis to be positive?
Also, assuming u, v, and w are vectors, what do the expressions u^v, v^w, and w^u mean? I don't know what it means to raise a vector to a vector power.

This is a alternative notation to "u x v" "v x w" etc {cross product}

 

[Mark44]

You didn't answer this question...

What does it mean for a basis to be positive?

This is a alternative notation to "u x v" "v x w" etc {cross product}

It's not one I've ever seen before, including in numerous linear algebra textbooks.

 

[member 587159]

This is a alternative notation to "u x v" "v x w" etc {cross product}

Where did you encounter such horrible notation?

 

[LCSphysicist]

Where did you encounter such horrible notation?

So probably you will not like to see it kkk:

[Boulos geometria analitica]

[Wikipedia]

And go on... Is not the first time that someone here don't know this notation, i will try to not use anymore

 

[LCSphysicist]

You didn't answer this question...
It's not one I've ever seen before, including in numerous linear algebra textbooks.

Well, a positive basis... Is just a way to adopt a orientation to the space, as well we do to a line or a plane.

Be a straight line, we can say the vector v is the positive orientation, so we have a set of:
all the vector with the same direction of v, and all the others opposites.

Be a plane, we can say: there is a pair of vector (v,u) , if we route v to u using the smaller angle, and this rotation resulting to be counterclockwise, so we can say that there is a Set in R2 (when we say about pairs) of vectors that route in counterclockwise, let's assume this positive, and the others are negatives.

Be now a space: If we have two basis in R3, and we can route one basis to superimpose on the other, in such way that the all vectors of the first basis always remains LI, so we have a positive basis, otherwise, a negative basis.

In physics language, we can guide the basis by dextrogira and levogira.

Say that both basis are positive, is say "that the basis have the same orientation or not "

" Using the cross product requires the handedness of the coordinate system to be taken into account (as explicit in the definition above). If a left-handed coordinate system is used, the direction of the vector n is given by the left-hand rule and points in the opposite direction. " {wikipedia}

Apparently outside Brazil you say as left handed coordinates and right handed coordinates?

 

[LCSphysicist]

I will question by another way so.
"Be (u,v,w) a basis fixing an orientation in space, so (u x v , v x w , w x u) has the same orientation. Prove it"

 

[etotheipi]

What does it mean for a basis to be positive?

If a basis $\beta = \{v_1, v_2, v_3 \}$ is "positive", then $v_1 \times v_2 \cdot v_3 > 0$. In other words, the basis is right-handed. At least I believe that's the convention.

 

[LCSphysicist]

If a basis $\beta = \{v_1, v_2, v_3 \}$ is "positive", then $v_1 \times v_2 \cdot v_3 > 0$. In other words, the basis is right-handed. At least I believe that's the convention.

In the case here i think this is not a good way to define, just make sense say "positive basis" if we compare with another adopted as a pattern. As said in the question "If ... is positive, so... is positive"
i could say " if ... is negative, so... is negative" too
I will post another question about this to see if there is another interpretation

But your sight is almost right, if we are comparing the basis ordinary i,j,k (adopting ijk positive) with this basis, we can find the changing of basis matrix and see if the determinant is positive or not.
$\beta = \{v_1, v_2, v_3 \}$
> 0

 

[LCSphysicist]

"Show the basis has opposite orientation {it could be said: show if the basis E is positive, so F is negative}"

 

[etotheipi]

View attachment 266403
"Show the basis has opposite orientation {it could be said: show if the basis E is positive, so F is negative}"

Well, suppose $\vec{e}_1 \times \vec{e}_2 \cdot \vec{e}_3 = k$, then$$-\vec{e}_1 \times \vec{e}_2 \cdot \vec{e}_3 = - k$$If one is a right handed basis, the other is left handed. This is fairly intuitive, since reversing a single vector in your basis reverses the handedness.

 

[atyy]

It's not one I've ever seen before, including in numerous linear algebra textbooks.

Where did you encounter such horrible notation?

It's the wedge product https://en.wikipedia.org/wiki/Exterior_algebra, just not very well typeset

 

[Infrared]

I think the argument in the OP is fine, as long as why you explain why the constants $\alpha,\beta,\gamma$ being positive follows from the original basis being positive.

 

[LCSphysicist]

I think the argument in the OP is fine, as long as why you explain why the constants $\alpha,\beta,\gamma$ being positive follows from the original basis being positive.

that is the main problem, i thought this intuitively, but i am still trying to know how to prove

 

[fresh_42]

that is the main problem, i thought this intuitively, but i am still trying to know how to prove

Here's a coordinate free solution I was thinking of:

You can read $(u,v,w)$ as a basis of $\mathfrak{sl}(2)$ and interpret $a \times b = [a,b]$. Then show that $u\longmapsto [v,w]\, , \,v\longmapsto [w,u]\, , \,w\longmapsto [u,v]$ defines a homomorphism and no anti-homomorphism.

 

[Infrared]

Since $u,v,w$ is positively oriented, $\det(u,v,w)=u\times v \cdot w=\beta w\cdot w$ must be positive, so $\beta>0$.

@fresh_42 Do you mean to use $\mathfrak{so}(3)$ instead of $\mathfrak{sl}(2)$? The Lie algebra $(\mathbb{R}^3,\times)$ is isomorphic to the former and not the latter.

 

[fresh_42]

Since $u,v,w$ is positively oriented, $\det(u,v,w)=u\times v \cdot w=\beta w\cdot w$ must be positive, so $\beta>0$.

@fresh_42 Do you mean to use $\mathfrak{so}(3)$ instead of $\mathfrak{sl}(2)$? The Lie algebra $(\mathbb{R}^3,\times)$ is isomorphic to the former and not the latter.

There is only one simple Lie algebra over $\mathbb{C}$ of dimension three, but $so(3)$ is probably better suited in this case.

 

[WWGD]

Well, a positive basis... Is just a way to adopt a orientation to the space, as well we do to a line or a plane.

Be a straight line, we can say the vector v is the positive orientation, so we have a set of:
all the vector with the same direction of v, and all the others opposites.

Be a plane, we can say: there is a pair of vector (v,u) , if we route v to u using the smaller angle, and this rotation resulting to be counterclockwise, so we can say that there is a Set in R2 (when we say about pairs) of vectors that route in counterclockwise, let's assume this positive, and the others are negatives.

Be now a space: If we have two basis in R3, and we can route one basis to superimpose on the other, in such way that the all vectors of the first basis always remains LI, so we have a positive basis, otherwise, a negative basis.

In physics language, we can guide the basis by dextrogira and levogira.

Say that both basis are positive, is say "that the basis have the same orientation or not "

" Using the cross product requires the handedness of the coordinate system to be taken into account (as explicit in the definition above). If a left-handed coordinate system is used, the direction of the vector n is given by the left-hand rule and points in the opposite direction. " {wikipedia}

Apparently outside Brazil you say as left handed coordinates and right handed coordinates?

I see, so you select one basis to be positive and other bases are pisutive too if the cgange-of-basis matrix has positive determinant?

 

[LCSphysicist]

I see, so you select one basis to be positive and other bases are pisutive too if the cgange-of-basis matrix has positive determinant?

yes, actually... I can't demonstrate this, but the book i learned it said:
"Enviamos o leitor interessado ao Capitulo 11, $ 10, do livro Introduction to Modern Algebra and Matrix Theory, cujos autores são O. Schreier e E. Sperne"

"To all readers interesting, we recommend to go to the chapter 11, $ 10, Introduction to Modern Algebra and Matrix Theory, authors: O. Schreier e E. Sperne"

I think the things become clear if someone share this part of book with us.