Видео

Hi abhiraja, thank you for your question. Indeed, in order to show that p vectors are indepedent, we have to look at the equation c1v1+...+cpvp= 0 and we have to show that the only solution of this vector equation is c1=0, c2=0,c3=0...cp=0. This is by definition what indepedence means. In this case we do this as follows; first we multiply with (A-lambda I)^{p-1} and this yields cp (A-lambda I)^{p-1} vp =0 , hence cp=0. Then we can repeat this argument with (A-lambda I)^{-p-2} and so on to show that all the other coefficients are zero. So the only solution of the vector equation c1v1+...cpvp = 0 is c1=c2=..cp=0 which shows that the vectors v1..vp are independent. Hope this helps a bit, good luck!

😮

Hi Ayishas804, thank you for your question. This is for a general matrix A, so in general you will not have v2 = e1. In the next videos in this playlist you can find an explicit example how you can find these generalized eigenvectors. Hope this helps.

@MatheMagician thank you sir for your reply,I am from India ❤️

No worries!

Brother this is from 7 years ago 😭 why bother telling him. Also it’s not difficult to understand just focus. Or move onto another video

Hi plankalkulcompiler, thank you for your kind reaction. On the older videos the volume is indeed too low. In the meantime we have improved this, so I hope this is better now in the more recent videos.

Hi dsgarden, thank you for your remarkt. You are right, I now added a flag to another video that explains how you can find the singular values; hope this helps!

Hi adithkiranmumar, thanks for your question. We are looking for a third vector orthogonal to u1 and u2. In order to find one we can start with any vector in R3 that is not in Span{u1,u2}. So I just picked an easy vector (e3). Hope this helps.

Hi, thank you for your question. I am not sure I understand exactly what you mean, but let me try to help. In this video we are actually solving two problems. First the SVD for A, which requires us to find U. And then we determine the SVD of another matrix B, which requires us to find another matrix U. Was this the problem?

@@MatheMagician I did not understand why we needed to find u3 for The Matrix B. I tried to test my answer with U(2x2) and it was perfectly true.

@@alibkhash Thank you for your question. Then you might be doing some other type of decomposition. In the SVD we explicitly impose the sizes of the matrices: we want U and V to be orthogonal matrices (hence square matrices), which means that we need U 3 by 3, V 2 by 2 and Sigma 3 by 2 in order to get B=USigmaV^T. You can generalize this idea to non-square U and V, but that is not what we did in this example. Hope this helps!

Hi Leon, thank you for your question. No, it is easier, it is becaucse (-1)^2=1: 1/(c-s)^2= 1/(-(-c+s))^2=1/(-1)^2(s-c)^2=1/(s-c)^2. Hope this helps.

@@MatheMagician Ahh, of course. Thank you

Hi OleJoe, thank you for your question. The branch cut has to connect 0 and infinity, the two branch points, so we do have a choice here. We want the positive real axis in our contour, because eventually we have to have the integral from 0 to infinity. We then also need to close the contour in some way, so you could close along the positive y-axis, but then you get some other integral along L2 (positive y axis instead of negative real axis) that we might not know. So maybe this works, but it will not make it easier as far as I can see. Hope this helps, good luck!

not true

The actual lecture quality was good, but the audio itself could use improvement

Hi AnmolKumar, thank you for your question. It should be: the total number of vectors in all cycles equals its algebraic multiplicity. So in this case we have 2 + 1 = 3 vectors in total, which indeed equals the algebraic multiplicity. I will check whether this is correct/clear on the other videos . Thank you for pointing this out and I hope this answers your question. Good luck!