I love the additional cool effects and the simple story revealing a cool mathematics feature. Keep up the great work

there is one thing that doesn't add up to me: this technique for calculating a function of a matrix is ​​based on the MacLaurin series expansion. But ln(x) is not defined at 0 and, in any case, if another development center is chosen, it will have a finite radius of convergence and the n-th term of the development does not contain x^n but (x-x0)^n .

@user-pr6ed3ri2k

Жыл бұрын

Apparently you can fix this by replacing 1 in the Taylor series of ln(x) at 1 with the identity matrix since that's basically 1 anyways

@VideoFusco

Жыл бұрын

@@user-pr6ed3ri2k All of the previous matrix function videos, which Dr. Peyam refers to to skip the intermediate steps, use the expansion around x=0. If he now uses an expansion around another point, he should demonstrate that this still allows the action of f to be transferred integrally to the diagonal matrix (especially, as in this case, if the series of f has a limited radius of convergence). This is not so obvious.

I like your candidness at the end.

So enthusiastic it is infectious. Thank you for sharing.

Always love your Linear Algebra videos❤

Great job, Dr. Peyam!

I love all your vids but especially linear algebra.

Simple and sweet.👌🏻

What is it useful for is a good question. I often use exp(A*dt) to convert continuous to discrete time matrices for simulation.

Interestingly, there is a relation for log in terms of power series, log(A) = - Σ (1 - A)^k / k, if it converges. With real (or complex) numbers a, the requirement for convergence is |a-1| edit: the spectral radius (maximum |eigenvalue|) must be < 1

@drpeyam

Жыл бұрын

Still true if you require ||A|| < 1 depending on which norm you use

Its last a video but benefit for renewing infromation 👍 Thank u doctor

it's useful for making us happy!

I'm probably butchering the terms below. So I usually see derivations of the diagonizability of certain functions done by using a Taylor series expansion. However the radius of convergence for the series representation of log is finite. Does that mean that using log in the case of a Matrix has to be within some radius of convergence in order for this to work?

fun and nostalgic. ln and exp are used in state space.

Can't you use the ln power series once you diagonilize the matrix? Is there a proof that the inverse function of the power series of e^x = the power series for lnx, or does the discontinuity at x = 0 exclude it as a metamorphic function?

So cool!

will this procedure work if we use quaternions instead of real numbers?

thx a lot

amazing!

Peyam can you show cool stuff of generating functions? It has great amounts of conteny

Could you do a video on the hyperdetetminant of a hypermatrix ? And maybe 2x2 for example

Intuitively I would think that the log of a matrix is defined via its power series. However, there is no power series for log(x), only for log(1+x). So, it would make sense to me, to take away the identity matrix away from the matrix in question and use that in the power series. From there, you can do the usual diagonalisation as was done in the video, and obtain the log of the matrix in question. Simply doing that for the matrix straight away, seems wrong.

Does log of matrix interpolate rotations, like SLERP for quaternions…..idk it’s been a while.

ln(A) makes sense because of the power series. If you compute the series A + A^2/2 + A^3/3 + ... you will get ln(I + A) as long as the series converges. Without this information, ln of a matrix looks so arbitrary and useless. But viewing ln(A) or exp(A) or sin(A) or whatever smooth function you want as a short-hand notation for a series expansion of a specific square matrix helps a lot. It's also consistent with regular notation since scalars are 1 by 1 matrices.

We can calculate logarithm by integration ln X = ∫_{ I }^{ X } T^{ - 1 }dT. We choose integral path as T = ( 1 - t )I + tX dT = ( X - I )dt Then we can express ln X as ln X = ∫_{ 0 }^{ 1 } ( ( 1 - t )I + tX )^{ - 1 }( X - I )dt

Very nice🥰🥰

Thanks a lot for this nice video👌

@drpeyam

Жыл бұрын

Most welcome 😊

1:55 could you please explain why taking ln of the matrix on the left means you take ln of each element in the diagonal matrix?

@IdunDied

Жыл бұрын

it's because it's the inverse of e^(Matrix). When you take e to the power of matrix, and do the diagonalized matrix form with eigenvalues diagonal in middle, and eigenvector matrices on each side, you take e to the power of each of the elements in diagonal matrix part and leave the eigenvector matrices on the outsides. So the inverse of that is to take ln of the elements in the diagonal matrix. Explanation of why that works is in dr peyam "e to a matrix" video. It's due to power series definition of exponential. TLDR is it's a sum and the eigenvector matrices multiply each term but aren't affected by the value of n you are on in the sum so can be taken out as constants. And then how the exponential power series (sum) works on diagonal(eigenvector) matrix is same as taking the exponential of the diagonal elements individually. So when you do inverse of that you do inverse exponential (ln) of each element.

Great video. Why the Ln function applied to the eign value matrix?

@sauzerfenicedinanto

Жыл бұрын

It is a consequence of Cayley Hamilton's theorem applied to matrix functions en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem

great work! I also want to know what if the matrix is not diagonalizable? what condition is necessary for logarithm of a non-diagonalizable matrix? it confused me for a while.

@drpeyam

Жыл бұрын

You can always use Jordan form 😁 And it works as long as the eigenvalues are >= 1

@jayyan4210

Жыл бұрын

@@drpeyam thank you!

so the e(A) is a power series coz I'm wondering what is the point of an exponential matrix ? are they're physics applications ? saludos mi amigo favorito ! 😅

Woohooo 🎉

But what if the matrix has at least one negative eigenvalue (non positive definite) ?

@drpeyam

Жыл бұрын

Imaginary!

@SpreeAlex

Жыл бұрын

@@drpeyam For all x

Dr Peyam, can you further explain what a matrix is and how does it work? I love your videos so much

@ymchen362

Жыл бұрын

Btw could you please make a video about "cos(x) in Deutsch" XD

@colindickson6099

Жыл бұрын

check out the linear algebra series on the channel 3Blue1Brown

@drpeyam

Жыл бұрын

See playlists

@ymchen362

Жыл бұрын

@@drpeyam 🙏

Are you alright man? We missed you lots!

@drpeyam

Жыл бұрын

Everything is well!!! Just tons of work, that’s all!

Ln2

What can I say !!? Cool ......

Why can't you just calculate the ln of each entry of the matrix right from the start? ln ((4 -2) (3 -1)) = ((ln 4 ln -2) (ln 3 ln -1))... the only problem is that the ln of negative numbers doesn't exist.

Is this Calculus?

@drpeyam

3 ай бұрын

Linear algebra

いいね！

First!

It is usefule rotation matrix in quantum mechanics, expressing e rather than ln

I don't think I was supposed to see this given I'm just in 12th standard preparing for jee 💀💀💀

@drpeyam

2 ай бұрын

👍