You can do that? logarithm of a rotation
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- เผยแพร่เมื่อ 4 ต.ค. 2024
- logarithm of a rotation. We calculate ln of a rotation matrix and see a really cool result from linear algebra. We then generalize this to 3 dimensions with Lie groups and so 3 and using metrics of matrices and Rodriguez formula. This is a must see for any calculus and linear algebra student! Enjoy this math adventure
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I immediately thought of complex numbers - that rotation looks like multiplication by i, so the logarithm is just ln(i) = pi/2 * i. And this looks just like the matrix at the end!
ah yes my ant math-brain managed to think of this too. awesome after 12 years after university math
Almost exactly what I thought. The rotation matrix represents a clockwise rotation, so I came up with multiplication by -i, and Log(-i) = - i pi/2 (where I use Log(z) to mean the principal branch of the complex logarithm).
if you replace matrices with complex numbers,
the result at 8:49 actually looks like
log( cos(α) + i sin(α) ) = (α+2πm)i
I did not expect the extra goodies at the end, super cool connections. Thank you for another interesting video
Rodriguez formula? I definitely used that in my robotics class at university. We coded forward kinematics, inverse kinematics, control algorithms, and path finding algorithms for a robot arm in our lab. We let Matlab do the heavy mathematical lifting though.
How coooool!!!
this approach is standard in control/guidance etc because it doesn't have degenerate coordinates like Euler's angles. I also wondered why such a super genius as Euler chose such a weak implementation of rotations, until I coded it up: you don't need a computer to find the inverse rotation if it's written in Euler's angles, you just read from right to left.
I really love how passionate and enthusiastic you are about mathematics, thank you for your videos ! :)
Suppose matrix [[0, -1], [1, 0]] corresponded by imaginary unit. Those, ln of rotation is also similar to ln of e^{it) (which is [[cost, -sint], [sint, cost]]). If I take ln of e^{it}, I get "i*t", which is also looks like matrix [[0, -1], [1, 0]], multiplied by angle "t" (if we use principled branch). Nice one :)
When you put it that way, it's obvious. I wonder if the other SO(n) correspond to exponentials of the "imaginary" part of some algebraic structure that can be modeled in nxn matrices.
You are simply awesome❤
feel nostalgic about chalk handprints on black shirt
More Lie Algebra content plz!!!
Hi Dr. Peyam!
What a miracle!
Very cool application 👍
That answer makes sense from the ln of a complex number with modulus 1, which is the argument of the number times the complex number with modulus 1 and argument pi/2. I think the same argument that the differential equation for exp gives circular motion when you plug in t times a right angle rotation applies to the matrix sort of rotation as to the complex plane multiplication sort of rotation, and ln being the inverse must extract "turn exactly left" times how far to turn.
I always think of complex numbers when I think of rotations, so this would be ln(e^{iπ/2}), but since i can be represented as a matrix I think these are equivalent
Just did the math, and if my math is correct then your final result is a matrix representation of iπ/2
anyone who's done quantum angular momentum homework already knows the eigenvalues of rotations. So here you have the spherical vector basis (not to be confused with spherical coordinates) in 2D....3D is a bit richer.
Nice :-) Wondered if the log gave the infinitesimal generator of all rotations and an small experiment in Maple confirms. m is your matrix logarithm, then with(LinearAlgebra):MatrixExponential(phi*m) gives a generic rotation. For phi=1 your starting point half pi rotation (90º) is recovered and for phi=4 would be 2pi radians rotation. Satisfying!
Awesome 👌
so cool!
THANK YOU!
Thank you very much.
MONOTONE FUNCTION ON R WHICH IS DISCONTINUOUS AT EVERY RATIONAL NUMBERS EXAMPLE EXPLANATION SIR PLEASE (NEXT VIDEO)
No
LOG(BASE = -2, VALUE = -8) ---> (1)
can be written as
= LN(-8)/LN(-2)
we know that LN(-K) = LN(K) + i(pi + 2N(pi))
N belongs to Z -> INTEGERS
so,
= [LN(8) + i((pi) * (1 + 2N))]/[LN(2) + i((pi) * (1 + 2M))] ---> (2)
for M, N belongs to Z
but simple overview of eqn(1) says it simply equals to +3
this case is possible in eqn(2) when both M & N is equal to (-1/2) which doesn't belongs to Z
so eqn(2) doesn't give real valued solution, but only complex solution, which is equivalent to the statement that eqn(1) has only complex solutions!
can you explain this inconsistancy sir
“but simple overview” is wrong
@@drpeyam but i meant is
(-2)^3=(-8)
Amazing
Fun fact i ≡ [ (0,-1) , (1,0) ] which means the ln(A)=ln(i)=iπ/2
love the way you say " + 2*me" (+ 2 peyam) (+2*pi*m)
It's funny that if you think of rotations as a transformation represented by e^Matrix, the natural logarithm is just the matrix corresponding to any given rotation. Havent watched the video yet, but that's my intuition.
Watch it :)
Wow, Dr. Peyam, your video was truly eye-opening! I knew a bit about the subject, but your explanation showed me much more to learn. It's like realizing math isn't just about adding and subtracting but involves a whole universe of concepts.
When we dive into higher dimensions, we can see that the Special Orthogonal Group, or SO(n), comprises n x n orthogonal matrices that have a determinant of 1 and correspond to rotations in n-dimensional space while maintaining vector lengths and angles.
Though it's possible to find the natural logarithm of an n-dimensional rotation matrix using methods like eigenvalues, eigenvectors, and diagonalization, the complexity increases as the number of dimensions grows. In such cases, Rodriguez's formula doesn't directly apply. Instead, we can look at the exponential map, which links the Lie algebra of a Lie group (in this case, the tangent space at the identity element) to the Lie group itself.
For the special orthogonal group SO(n), its Lie algebra, represented as so(n), comprises n x n anti-symmetric matrices. Given an anti-symmetric matrix A from so(n), we can compute the corresponding rotation matrix R in SO(n) by finding the matrix exponential of A (exp(A)). This can be achieved using methods like power series expansion. Conversely, to find the logarithm of a rotation matrix R in SO(n), we can compute the matrix logarithm of R (log(R)), which retrieves the original anti-symmetric matrix A from so(n).
I always like to try and push it and find it the most interesting to think of the infinite-dimensional case.
One approach is to use the hyperreal number method to discretely study the continuous group of invertible matrices that contains A. This involves constructing a hyperreal version of the group, a discrete approximation of the original continuous group. In the hyperreal version, if Lie theory and functional analysis permit, it might be possible to construct a sequence of finite-dimensional matrices that converge to A in some sense, then compute the logarithm of each matrix in the sequence using standard techniques. The limit of these logarithms can then be taken to obtain the logarithm of A.
Hi Peyam! Can you clear this up for me? It is not clear how you applied the natural log function or the definition of the natural log function on matrices. I see that you are applying it to a similar Diagonal matrix and that property holds for exponents but how does that hold for applying functions to a matrix? Are you treating the natural log function as a multiplication of a transformation matrix?
I know Dr. Peyam has done a few videos where he applied a function to a matrix and went through by expanding the function into a power series representation and then evaluated by substituting the diagonalized form P*D*P^-1 into the series. I found this one where he takes the arctan of a matrix: th-cam.com/video/_8DS4LRmwO4/w-d-xo.html
He explained it several times before. Basically, for any f(x) that can be expanded into a power series, you can use diagonalization to turn it into f(A), because
A=U D U^-1
A^n=U D U^-1 U D U^-1 ... U D U^-1 = U D D ... D U^-1 = U D^n U^-1
and also you get D^n by taking ^n of each diagonal element independently.
Then you expand f(A) into a power series, and you get
f(A)= f0 I + f1 A^1 + f2 A^2 ... = U [ f0 I + f1 D + f2 D^2 ...] U^-1
which, if you think about it for a while, simplifies to
f(A) = U f(D) U^-1
where you get f(D) by taking f(x) of each diagonal element independently.
Is the distance measured in the same sense as Euclidean distance?
Why is it enough to take the ln on the middle matrix?
At 4:41 would it be possible to take out the scalar factor of (pi/2 + 2pi m) from the central matrix and then bring that constant to the front? This would leave it as (pi/2 + 2pi m) multiplied by A (written in its diagonalised form) and so the inverse and multiplication of the matrices from 4:41 - 7:05 would not have to be calculated.
what if we use quaternions? Will it be spins in 4D?
I would look to what rotors are in geometric algebra (which goes further than 3D btw, and pretty dimension-agnostic) and why quaternions are a special case of them (and in physics they show up as the "pauli matrices", which also give rise to rotations in 3D).
I always find it scary when i see complex numbers. not sure how to build intuition behind it.
use them
Why when we have a sqwer root of a namper we just take the posetive solom
Hello Sir can you make a brief video on limit point in topology?
because we all know in real analysis that infinite set has limit point(s), but how is it possible in topology of finite subset? I hope you understand what I want to say.
There’s a topology playlist
So is SO3 the collection of objects that transform like rotation matrices?
Once you had complex eigenvectors in the decomposition, I expected to see a conjugate of the inverse. Obviously not needed. Is that because the original matrix A was real? ( excellent vid for a saturday morning!)
Excuse me, Dr Peyam, but how do you know that the additional i*2*pi*m values for the natural logarithms of the eigenvalues have the same value for each of the dimensions?
Shouldn't that m (or n) not be an arbitrary integer each time one takes the natural logarithm in the complex plane, and henceforth don't need to be equal for the different coordinates and when they're not equal can't be pulled out of the matrix?
That follows from the fact that exp(2pi im + x) = exp(x)
@@drpeyam @Dr Peyam Which follows from Euler's formula, just like exp(2pi i)=1 and I already know that.
I do see that if x=i and y=-i then ln(x)=i*pi(1/2 +2*m) and ln(y)=i*pi(-1/2+2*n) with n,m being integers but not why also n=m.
my ii
😆😆😆 "determinant of Muhammad Ali", "multiples of myself" what else is coming?
Dear Dr. Peyam, I always follow you with great interest. I wanted to offer you a new field of investigation. Remember your videos when you computed the half derivative of x^n? Well, let's say that that was a generalization of the concept of derivative. I want to propose the following problem: Let us consider the arithmetic operations that we classify according to these steps:
1) sum
2) multiplication
3) exponential
4) tetration
5) pentation
6)... and so on
let us now consider a very simple example where we gradually apply the various steps above; for simplicity we use a numerical example with the number 3:
1) 3+3=6
2) 3x3=9
3) 3↑3 =27
4) 3↑↑3 =...a very high number!!!
5) 3↑↑↑3 =...an even higher number!!!
now let's establish a graphic symbol that denotes the arithmetic operation we are referring to, for example, we can use the following symbol: ┌ n ┘ (obviously one can invent the symbol he likes best), at this point, let's generalize the writing some arithmetic operations like this:
1) 3 ┌ 1 ┘3 = 6
2) 3 ┌ 2 ┘ 3 = 9
3) 3 ┌ 3 ┘ 3 =27
4) 3 ┌ 4 ┘ 3 =...a very high number!!!
5) 3 ┌ 5 ┘ 3 =...an even higher number!!!
Well, at this point the game is to create a "generalized" arithmetic operation that admits any value of n, therefore not necessarily a natural number, but also a fractional, irrational, imaginary, complex number, a matrix ...
For example, the first step would be to define: ┌ 1.5 ┘that is, what value it assumes:
3 ┌ 1.5 ┘ 3 = ???
certainly, as written in steps 1) and 2), we can state that:
6 < (3┌ 1.5 ┘3) < 9
It would be nice, then, to construct this generalized arithmetic operation, I'm sure, which would open up a whole new field of mathematics....
:)
First!