Could you please explain what you wrote at 17:09 ? How'd you get all zeroes in the diagonal of the matrices? That can only happen when A has only 2s in its diagonal (and in that case it's already diagonizable), but we don't know that, do we
He used the rank-nullity theorem - the dimension of the image plus the dimension of the kernel is the size of the matrix (it should be clear dim(Im) (i.e. the rank of A) is 1)
@@TheDelcin actually it's too much for beginners: he doesn't explain the theorem itself, WHY can we just have ones on the superdiagonal and zeroes everywhere else. He just quickly described what Jordan form looks like
The only well-explained & sufficient video about Jordan form on youtube. Please, keep making more videos
i don't understand why you didn't explain exactly how those 1s are determined.
At the moment, this is my favourite video on youtube.
Excellent!
This was a perfect lin. alg. refresher for my controls systems course!
pretty good; but shouldnt you have mentioned the correlation of selecting jordan blocks based on its dependancy on minimal polynomials?
This is my last hope how to pass my linear alg. exams. Thanks !
Perfectly explained, thanks a lot!
Master class teaching right here.
why didnt you work an entire example...
Thank you for a well explained video where the audio is clear
nice one.Enjoyed and useful
10:17 It seems S matrix is incorrect. Performing S^-1 * A * S doesn't yield the result you've shown.
super nice video
Could you please explain what you wrote at 17:09 ? How'd you get all zeroes in the diagonal of the matrices? That can only happen when A has only 2s in its diagonal (and in that case it's already diagonizable), but we don't know that, do we
A has 2s in the diagonal for sure because its characteristic polynomial (with out knowing if A is recognizable or not)
At 5:02, i don't understand why the dimension of the kernel is 1. can you explain to me more, please?
He used the rank-nullity theorem - the dimension of the image plus the dimension of the kernel is the size of the matrix (it should be clear dim(Im) (i.e. the rank of A) is 1)
All the examples are incomplete , just what's the point from that .
Not for beginners obviously. Read Gilbert Strang
@@TheDelcin actually it's too much for beginners: he doesn't explain the theorem itself, WHY can we just have ones on the superdiagonal and zeroes everywhere else. He just quickly described what Jordan form looks like
very good
Very good explanation！ Thanks
perfectly explained!
well explained thank u
Thanks so much
nice video but the writing is not clean, sorry.