LU Decomposition
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- čas přidán 24. 07. 2018
- MIT 18.06SC Linear Algebra, Fall 2011
View the complete course: ocw.mit.edu/18-06SCF11
Instructor: Ben Harris
A teaching assistant works through a problem on LU decomposition.
License: Creative Commons BY-NC-SA
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Ben really did an amazing job I have been watching videos of the LU decomposition for 2 hours and your explanation is the best so far
absolutely enjoying it. understood the concept the moment i finished the video!
I appreciate your help. Now I have a better understanding of matrices multiplication through your precise description.
these videos are gold :)
Ben, you explained nicely..
Thank you
I can easily understand LU decomposition by following your process in a timely manner
No you dont
Nah @@lugia8888
@@lugia8888 dawg why you so salty lol
Thank you very much, definitely useful for me!
Dude you really nailed it!! First I didnot understand the actual procedure from lecture not I get it
and I was wondering why L and U decomposition would take less amount of steps to compute and now I have my answer thanks to your clear explaination
Am watching now thanks to Instructor Harris
Good review. I remember I was puzzled why L doesn't have minuses when it eliminated numbers.
love this course
Very Helpful! Quick and Clear!
He's pretty good. Thank you so much, Ben :) :)
Thank you! I think he explained it very well
Thank you man, it was a really clear explanation
Thanks this will be huge help
Thanks for your video
What about the case when we need row interchange?
Wait, is it just me who's finding his LU doesn't end up the original A? I was following all of the steps correctly until the last part where the LU is not equal to A.... Anyone else?
Yeah bro same in matrix U (u33 = 0 ) not a-b ,I used a different method got u33 as zero and A = LU
Some one maybe explain to me why " you can't find LU decomposition if you have to do a row exchange in elimination"(4:08) it seems to me that it is because in this problem we happen to have 0 in the right of a in row 2, if a is zero, then row two is dependent row. Without full pivot we can't do LU decomposition. But in other situation, even if we have row exchange in elimination step, we could still find LU decomposition but not in a straight forward way. Someone please help!
please explain the last comment you have made regarding permutation.. why if permutation does not allow LU decomposition
because if we would do permutations the resulting matricies wouldnt be triangular anymore. ->triangular matricies are our goal
(i know the comment is old, its just for anyone having the same question)
If we need to change the pivot can we not just do a permutation and added to the elimination
Then you won't get a LU decomposition.
U done it beautifully in L 21. There will be -a not just a 😅 8:40
i love his voice
Why we not take into account the row exchanges in LU decomposition. As we know that the elementary matrix corresponding to it is invertible(equal to its transpose). So why row exchanges are not considered?
But why would you want to exchange rows in this problem? All pivot positions are already non-zero.
thats a plu decomposition
I liked this one.
How do you get -b/a for Esub32 in the third elimination matrix?
We want to turn the b in row 3 to 0 using the pivot in row 2. In order to do this, the pivot, namely a needs to be first divided by a and then multiplied by b followed by a subtraction from row 3. Translating this into matrix form is E_32.
Is there any reason why the sequence is always E32->E31->E21?
I guess unless it will be hard to compute L=E21 -I × E31 -I × E32 -I , right?
because it is elimination, it is about row operations,so the matrix must be on the left of the A matrix, coz it is step by step, so it must be this order
There is a good reason why it's that exact order.
Having E32 x E31 x E21 x A = U, in order to end up only with A on the left side, we need to multiple on the left with a matrix that would result in I x A = L x U.
The definition of I is: I = A x A-1 or I = A-1 x A
To get I from E32 x E31 x E21 we need to multiply it on the left with the inverse of each matrix (in reverse order).
E21-1 x E31-1 x E32-1 x E32 x E31 x E21 = I, because E32-1 x E32 = I, and then E31-1 x I x E31 = I, and then E21-1 x I x E21 = I
Even E31->E32->E21 gives the same L in this case, just need to follow order of inverses to be multiplied(E21-1->E32-1->E31-1). Not sure if it's limited to this question only.
Can anyone explain me what the last part meant?
That a-b can be zero and singular matrices can have LU decomposition.
Singular matrices can have U, they just can’t be eliminated to I.
this is great
But the U matrix contains pivots and pivots can't be zero, so (a - b) can't be zero. Right?
idk
why a-b can be zero? Suppose a = 1 and b = 1, then the pivot will be zero
a was assumed to be zero because it is required to be used as a pivot in order to obtain u. i see what you mean by your question because a-b is the next natural pivot location, but it isn’t necessary for a-b to actually be a pivot since we aren’t eliminating anything underneath of its location. we can still construct an upper triangular matrix even if none of the elements on the diagonal are valid pivots. all that matters is that there are zeros below the diagonal.
i know this is 8 months late but this is for anyone who read your comment and is still curious.
@@logancastaway2064 I really curious about that and I have searched comments to know that. Thank you so much!!
Singular matrices are matrices that doesn't have inverse.
Thanks. liked.
nicely
I still don't get it. I think the e matrices are tripping me up.
You're BEN
he is explaining it so well i can even understand and i am a 11 year old
👍
Looking forward to hearing about your Fields medal sometime in the next 29 years
@@steve6012 XD
It seems that he left-multiplied each successive matrix by the eliminators, rather than right-multiplying!
Hey can I get into MIT?
See mitadmissions.org/ for info.
From professor Strang, when the pivot is ZERO, it can be multiplied by a permutation matrix, to change the the row, so PA=LU, it should be ok, why a =0 is not discussed?
Check the next lesson. ;)
Challenge: Take a shot everytime he says "Good".
11 shots...
Wait... as the second row is divided by b/a which means b/a is not zero, that leads to a≠0 and b≠0...
second row times b/a, not divided by b/a
omnitrix waiting for you mate
Hello Ben
Wouldn't it be necessary to highlight that "a - b" must be non-zero?
He mentions it at the very end of the video. I believe it is alright for a-b to be 0 since the location of the entry does not require a row exchange which would prevent us from creating an LU decomposition
1) Check that your solution is valid. Check that LU gives you A. Just like with checking solutions of differential equations.
2) the a-b pivot in matrix U can be 0, because we don’t have to do a row exchange to get U.
That’s the only time when we can’t do LU decomposition.
In particular, singular matrices can have LU decompositions.
The 2nd point makes no sense to me.
The second point is true enough, up to a certain point. You can't perform an LU decomposition for some matrix A which needs a row exchange; instead, you have to reorder the matrix A into some other matrix PA (by means of a permutation matrix) such that PA can be factored into LU.
Singular matrices can have LU decompositions; in fact, singular matrices can have multiple LU decompositions.
I miss blackboards. They’re the OG boards
Güd
I think a=0 b=0 can be the solution because A = IA is also LU decomposition
No it can't be. You essentially said L=I & U=A, which can not be true, as you need to perform some set of elementary operations to reach A (as shown in the video).
Hope it was helpful!
Ah, good point. Technically, if a & b = 0 -> A=U & L=I, so A=IA is a solution.
What you said is called the "trivial" solution, matrix A should always be equal to IA which is also equal to AI. and you can call this what ever you want, not only LU. I guess by now you have passed the course ! congrats =)
@@endogeneticgenetics a = b = 0 has nothing to do with whether A = IA holds or not. A = IA is always true.