Thank you so much for these clearly-presented examples! Recently I was tutoring a student in Linear Algebra who just started the Jordan Canonical Form, which I'd never actually done myself. But I was able to learn and understand it well enough from these videos. Thanks again!
@@Farawayaz pues mi maestra es latina, no sabe ni lo que esta leyendo o donde esta parada y quitaron al coordinador de ingenieria para darle esa clase a ella, asi que desearia esa dicha de que me perdiera cuando escriben un numero jajja exelente video
Could I get the canonical "A" matrix form ("J") directly by Jordan's blocks? I consider the main diagonal of the J matrix and put all the eigenvalues; geometric multiplicity are jordan's blocks. So i don't need the "S" base change matrix. Thanks for the reply.
Since the geometric multiplicity for eigenvalue 2 is not equal to the algebraic multiplicity, we need to find a generalized eigenvector for eigenvalue 2.
The eigen space of the v2 is the span of (1,0,0) but it dosent change the outcome, and the zero vector is a trivial solution anyway, putting the zero vector wouldnt be right
@@LYNXzTwist What are you talking about? v3 can, indeed, be (1,1,1). So the following matrix (in matlab notation) s = [1 1 1;0 1 0;0 1 1] does indeed put A in Jordan form just as well.
I guess that you don't need the answer anymore, but I'll answer anyway for the future viewers. The video is literally about that. Did you mean characteristic equation? You should calculate det (A - x*E), where x is unknown (will be eigenvalue) and E is identity matrix.
It's okay. But this didn't really work for me. Try doing this with this simple matrix: [2, 0, 1; 0, 2, 0; 0, 0, 2]. You'll end up with v1, v2, v3 AND v4. You'll get 4 vectors and then you're not able to create the S you showed in the video...
@@ajsdoa6282 So first of all the matrix [2, 0, 1; 0, 2, 0; 0, 0, 2] makes it actually really easy to compute since our characteristic polynomial just looks like this: pa(x) = (x-2)^3 then just apply the formula used to calculate the eigenspaces... You will not end up with 4 eigenvectors since the root of your polynomial can only be 3 not higher. Also.. the 1 is just a normal way to write it... it's quite common in europe to write it that way. :)
Thank you so much for these clearly-presented examples! Recently I was tutoring a student in Linear Algebra who just started the Jordan Canonical Form, which I'd never actually done myself. But I was able to learn and understand it well enough from these videos. Thanks again!
Mrs. Linda, You are Great!
That is the weirdest drawn “ 1” I have ever seen
Rest of the video was brilliant
its actually how Germans write the number. My German prof always did it and we were so lost :')
@@Farawayaz I do agree! I'm experiencing this too in Germany! :D
#Germanway
@@abdullahalimran4416 I probably am about to I guess.
@@Farawayaz pues mi maestra es latina, no sabe ni lo que esta leyendo o donde esta parada y quitaron al coordinador de ingenieria para darle esa clase a ella, asi que desearia esa dicha de que me perdiera cuando escriben un numero jajja exelente video
Thank you for helping me pass my Linear algebra 2 exam!!! :D
Im Sitting mine Tomorrow 😂
mine is in a few minutes I hope ir works
Mine is After 90 minutes 🙃🤐
@@ptwnight9326 goodluck!
Thank you, besides theoretical knowledge, I think I better first know an example, it helps me a lot 😃
Amazing explanation, thanks for the video!
Narek Grigoryan I see what you did here
Andreas Ierodiaconou Πες τα ρε Αντρέα, μας τρέλανε η τύπισσα
Could I get the canonical "A" matrix form ("J") directly by Jordan's blocks? I consider the main diagonal of the J matrix and put all the eigenvalues; geometric multiplicity are jordan's blocks. So i don't need the "S" base change matrix. Thanks for the reply.
So clear and helpful. Thank you
Excellent oresentation.only should tell what method be adopted if S not invertible
Why (A-2E)v3 us equal to v2?please need explanation
Since the geometric multiplicity for eigenvalue 2 is not equal to the algebraic multiplicity, we need to find a generalized eigenvector for eigenvalue 2.
Because the eigenvalues have multiplicity
why there is a zero for v3 and one for v2 in the first element of the v3 and v2 vectors respectively?
Maam, can you explain this please
The eigen space of the v2 is the span of (1,0,0) but it dosent change the outcome, and the zero vector is a trivial solution anyway, putting the zero vector wouldnt be right
And the reason why v3 is not (1,1,1) is because it dosent make it trivial or not eithee way, and the root space of v2, v3 spans (1,1,1) anywya
@@LYNXzTwist What are you talking about? v3 can, indeed, be (1,1,1). So the following matrix (in matlab notation) s = [1 1 1;0 1 0;0 1 1] does indeed put A in Jordan form just as well.
why did you write 1 in that way...
Thank u so much 🥰
The way the ones are drawn make this video so much harder to follow for me. So bizarre
you could have done better explaining how it works, you just wrote down on the board the solution. Thank you tho.
how she has calculated the eigen vector?
I guess that you don't need the answer anymore, but I'll answer anyway for the future viewers. The video is literally about that. Did you mean characteristic equation? You should calculate det (A - x*E), where x is unknown (will be eigenvalue) and E is identity matrix.
first balabalab than magic than more magic here we get the solution do you understand?
Strange "1".
sehr gut :)
Thank you Linda :)
Very good!
Thank youuuuu 😭😍
isn't v3 a 2 dimensional solution space of ((0 1 1),(1 0 0))???
Your 1:s looks like V:s upside down
It's okay. But this didn't really work for me. Try doing this with this simple matrix:
[2, 0, 1;
0, 2, 0;
0, 0, 2].
You'll end up with v1, v2, v3 AND v4. You'll get 4 vectors and then you're not able to create the S you showed in the video...
@@ajsdoa6282 So first of all the matrix [2, 0, 1;
0, 2, 0; 0, 0, 2] makes it actually really easy to compute since our characteristic polynomial just looks like this: pa(x) = (x-2)^3 then just apply the formula used to calculate the eigenspaces...
You will not end up with 4 eigenvectors since the root of your polynomial can only be 3 not higher.
Also.. the 1 is just a normal way to write it... it's quite common in europe to write it that way. :)
I understand, actually