Graduate Lie Bracket * Matrix * vector (Need proof)

[Trying2Learn]

TL;DR

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?

 

[fresh_42]

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.

 

[Trying2Learn]

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.

 

[Trying2Learn]

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.

 

[fresh_42]

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.

 

[Trying2Learn]

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?

 

[fresh_42]

Well, at least you are right, it is zero:

https://www.symbolab.com/solver/mat...x}\cdot\begin{pmatrix}x\\ y\\ z\end{pmatrix}=

But if you do not want it in coordinates, then the calculation laws of cross- and dot product is probably the best you can do, i.e. use the rotation operator.

 

[Trying2Learn]

Well, at least you are right, it is zero:

https://www.symbolab.com/solver/matrix-calculator/\begin{pmatrix}0&-c&b\\ c&0&-a\\ -b&a&0\end{pmatrix}\cdot\begin{pmatrix}0&-z&y\\ z&0&-x\\ -y&x&0\end{pmatrix}\cdot\begin{pmatrix}0&-c&b\\ c&0&-a\\ -b&a&0\end{pmatrix}\cdot\begin{pmatrix}x\\ y\\ z\end{pmatrix}-\begin{pmatrix}0&-z&y\\ z&0&-x\\ -y&x&0\end{pmatrix}\cdot\begin{pmatrix}0&-c&b\\ c&0&-a\\ -b&a&0\end{pmatrix}\cdot\begin{pmatrix}0&-c&b\\ c&0&-a\\ -b&a&0\end{pmatrix}\cdot\begin{pmatrix}x\\ y\\ z\end{pmatrix}=

But if you do not want it in coordinates, then the calculation laws of cross- and dot product is probably the best you can do, i.e. use the rotation operator.

Thank you!

t

 

[fresh_42]

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 $W_1=-W_0^\tau\, , \,S_1=-S_0^\tau .$

 

[Trying2Learn]

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:

W_ij = -e_ijk w_k
S_ij = -e_ijk s_k

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