Lie Bracket * Matrix * vector (Need proof)

In summary, In summary, the symbolic manipulator in Matlab gives me the result of a vector that is perpendicular to both w and s. This means that the cross product of v and [W,S] is zero, as desired.
  • #1
Trying2Learn
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TL;DR Summary
Relationship of Skew Symmetry, vectors, cross product and Lie Bracket
As an aside, fresh_42 commented and I made an error in my post that is now fixed. His comment, below, is not valid (my fault), in that THIS post is now fixed.Assume s and w are components of vectors, both in the same frame

Assume S and W are skew symmetric matrices formed from the vector components

I want to prove this (that I CAN commute the first two terms on each side)

W*S*W*s = S*W*W*s

Thus, I want to prove

W*S*W*s - S*W*W*s = 0

(W*S-S*W)*W*s =0

However, the parenthetical is the Lie Bracket. Thus

[W,S]*W*s = 0

So, is the above, true?

I can show it works for any general skew symmetric system, using the symbolic manipulator in Matlab

But can someone explain why this is true?
 
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  • #2
You need to specify the dimensions of the vectors are taken from, or the dimension of the matrices. And what is ##s## for? Did you mean ##V=S## and what does ##W,V## consists of the components of ##v,w## mean?

If the dimension is greater than two, and ##s\neq \vec{0}##, and ##W## is invertible, then ##[W,S]\neq 0## in general, i.e. for arbitrary ##S##.

Maybe you can give an example because your question is too flawed to answer.
 
  • #3
fresh_42 said:
You need to specify the dimensions of the vectors are taken from, or the dimension of the matrices. And what is ##s## for? Did you mean ##V=S## and what does ##W,V## consists of the components of ##v,w## mean?

If the dimension is greater than two, and ##s\neq \vec{0}##, and ##W## is invertible, then ##[W,S]\neq 0## in general, i.e. for arbitrary ##S##.

Maybe you can give an example because your question is too flawed to answer.

The dimension of all is 3.
w is related to an angular velocity vector
s are the position vector

In the process, I derive this term that I KNOW must be zero (I just can't prove it):
W*S*W*s = S*W*W*s

I can REASON my way around it like this:

I want to prove

W*S*W*s - S*W*W*s = 0

Thus, I want to prove

(W*S-S*W)*W*s =0

However, the parenthetical is the Lie Bracket (A Commutation relationship). Thus

[W,S]*W*s = 0

OK, so v = W*s is a vector perpendicular to w and s

The Lie Bracket is a skew matrix representing a vector perpendicular to w and s

This means that [W,S] and v are parallel

Thus, their cross product must be zero

However I do not like that proof. It seems to be a lot of handwaving and too much reliance on the cross product.
 
  • #4
fresh_42 said:
You need to specify the dimensions of the vectors are taken from, or the dimension of the matrices. And what is ##s## for? Did you mean ##V=S## and what does ##W,V## consists of the components of ##v,w## mean?

If the dimension is greater than two, and ##s\neq \vec{0}##, and ##W## is invertible, then ##[W,S]\neq 0## in general, i.e. for arbitrary ##S##.

Maybe you can give an example because your question is too flawed to answer.
And I apologize: there was no need for v or V. I meant s and S.

I have corrected that mistake in my original post.
 
  • #5
I still have difficulties understanding it. If ##[W,S]Ws=0## and ##W## is regular, then ##[W,S]=0## but this isn't the case for arbitrary skewsymmetric matrices of dimension three. So either ##s\in \ker W## or ##Ws \in \ker [W,S].## Hence we need to look at your additional geometric conditions. However, I don't see how to apply them.
 
  • #6
fresh_42 said:
I still have difficulties understanding it. If ##[W,S]Ws=0## and ##W## is regular, then ##[W,S]=0## but this isn't the case for arbitrary skewsymmetric matrices of dimension three. So either ##s\in \ker W## or ##Ws \in \ker [W,S].## Hence we need to look at your additional geometric conditions. However, I don't see how to apply them.
I am sorry for confusing you: it is my fault for lack of precision (which actually derives directly from my very confusion)

So here... I wrote a Symbolic Manipulator Matlab script. This script gives me the result of ZERO.

I now wish to prove this

---------------------------------------------------Test.m-----------------------------------
% Declare symbolic varitables to be real
syms w1 w2 w3 s1 s2 s3 v1 v2 v3 real

% Create two symbolic vectors
w = [w1;w2;w3]
s = [s1;s2;s3]

% Create their associated skew symmetric form (as if to be used in the cross product)
W =[0,-w3,w2;w3,0,-w1;-w2,w1,0]
S =[0,-s3,s2;s3,0,-s1;-s2,s1,0]

%Perform the operation
T=simplify((expand(W*S*W*s-S*W*W*s)))

I get the components of a zero vector!
I get exactly what I want.
(This is from the realm of rigid body dynamics -- nothing advanced as you would find it relativity or advanced geometric space-time, etc.)
Why?
 
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  • #8
  • #9
Trying2Learn said:
Thank you!

t
Maybe you can shorten the calculationn by splitting ##W=W_0+W_1\, , \,S=S_0+S_1## into upper and lower triangular matrices and use the facts that they are multiplicative closed and ##W_1=-W_0^\tau\, , \,S_1=-S_0^\tau .##
 
  • #10
fresh_42 said:
Maybe you can shorten the calculationn by splitting ##W=W_0+W_1\, , \,S=S_0+S_1## into upper and lower triangular matrices and use the facts that they are multiplicative closed and ##W_1=-W_0^\tau\, , \,S_1=-S_0^\tau .##

Actually, I found a better way.

Remember, in my problem, s and w were the vectors
S and W were the corresponding skew symmetric forms.

We already know that Ws and Sw are manifestations of the cross product.
Thus:
Ws is perpendicular to w and s
Sw is perpendicular to w and s.

Next, I investigate:
[W,S] = WS - SW

I use an alternating tensor to represent a skew symmetric form.

I find any skew symmetric form has elements such that:

Wij = -eijk wk
Sij = -eijk sk

Then I go through the process of forming
[W,S]ij = (WS - SW)ij

I find that that last form is ALSO perpendicular to w and s.

Thus,
[W,S] * Ws represents a cross product of two parallel vectors
[W,S] * Sw represents a cross product of two parallel vectors[W,S]Ws = 0
[W,S]Sw = 0

And I am done
 
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FAQ: Lie Bracket * Matrix * vector (Need proof)

What is the Lie Bracket of a matrix and a vector?

The Lie Bracket of a matrix and a vector is a mathematical operation that combines the two objects to create a new vector. It is denoted by [A, v] where A is the matrix and v is the vector.

What is the proof of the Lie Bracket * Matrix * vector formula?

The proof of the Lie Bracket * Matrix * vector formula involves using the Jacobi identity, which states that the Lie Bracket of three objects is equal to the sum of the cyclic permutations of the three objects. By applying this identity to the Lie Bracket of a matrix and a vector, we can derive the formula [A, v] = Av - vA.

How is the Lie Bracket * Matrix * vector used in physics?

The Lie Bracket * Matrix * vector formula is used in physics to study the dynamics of systems with continuous symmetries. It is particularly useful in the field of quantum mechanics, where it is used to calculate the time evolution of quantum states.

Can the Lie Bracket * Matrix * vector be extended to higher dimensions?

Yes, the Lie Bracket * Matrix * vector formula can be extended to higher dimensions. In fact, it can be generalized to any Lie algebra, which is a mathematical structure that describes the symmetries of a system.

Are there any real-world applications of the Lie Bracket * Matrix * vector formula?

Yes, there are many real-world applications of the Lie Bracket * Matrix * vector formula. It is used in fields such as robotics, control theory, and computer graphics to study the dynamics of systems with continuous symmetries.

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