Question

linear algebra: A=Lu

Answer 1

http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/factorization-into-a-lu/MIT18_06SCF11_Ses1.4sum.pdf

Answer 2

\[EA=u\\A=E^{-1}u\]

is it guaranteed that \(E^{-1}\) is a lower triangular matrix? (if no row exchanges are made?)

i don't understand why we are factoring it this way,
does it help solve AX=b faster ?

Answer 3

under the section: how expensive is elimination
is talking about cost of finding A^-1 , or of factoring A as Lu, or of solving Ax=b
?

Answer 4

@ganeshie8 @phi

Answer 5

Hi

Answer 6

hey :)
finally changed ur pic i see
i do like ice cream :P

Answer 7

Haha that is just to give myself some motivation to eat

Answer 8

is it guaranteed that E−1 is a lower triangular matrix?

notice E is the identity matrix, with one additional value at location row i, col j
when we do elimination, we always eliminate entries in A below its diagonal (assuming square, and full rank)

Answer 9

that means all the E's are lower triangular.
And we can show that multiplying lower triangular matrices results in a lower triangular matrix.

Answer 10

hmm... i think i get that, i'll try with some examples later and see if i get it fully

Answer 11

so why are we doing this? how does it help with gauss elimination ?

Answer 12

Don't you appreciate that lower triangular matrix represents a set of equations that can be solved trivially ?

Answer 13

Look at below eqns
x1 = 2
x1 + x2 = 3
etc

Answer 14

yes, that was the previous lecture
\[AX=b\\(EA)X=Eb\]

EA is a lower triangluar matrix from which we can pretty much easily read off the values for the unknowns

i dont get why we are factoring A=Lu

Answer 15

I thought Lecture 2 goes into the details of how to solve Ax= b using LU factorization

Answer 16

lecture 2 does not discuss A=Lu

only E(A)=U
and how this makes solving for X easy

Answer 17

lecture 3 say that we can factorize A as Lu,
but dosen't seem to explain why we want to.
is it covered in subsequent lectures?

Answer 18

how is
\[AX=b\\(Lu)X=b\]

easier than
\[(EA)X=Eb\\UX=Eb\]

Answer 19

i haven't gone through the problems/ recitation video..
is it covered in those?

Answer 20

LU x = b
first you solve
L y = b
using forward substitution to find y
then you solve
Ux = y
using back substituion

Answer 21

whats 'y'
is this method covered later on?
is it computationally less expensive than the earlier method (EA)X=Eb ?

Answer 22

y is an intermediate result (a vector)
https://en.wikipedia.org/wiki/Triangular_matrix#Forward_and_back_substitution

Answer 23

thanks,
i've got to go now, please do post any thoughts you have.
i'll read that when i can and get back to you :)

Answer 24

in other words with
LU x = b

Ux will be a vector. call it y.

L y = b
we know L and b, so we find y
knowing
Ux = y
and knowing y and U, we solve for x

Answer 25

"E is the identity matrix, with one additional value at location row i, col j"
this is always true assuming a typical case?

Answer 26

here in the pdf
we have
\[E_{32}=\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & -5 & 1 \end{matrix}\right]\]

and
\[E_{32}^{-1}=\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 5 & 1 \end{matrix}\right]\]

as in, the inverse of the eliminaiton matrix is just swapping signs of elements below the diagonal

is that how Prof. Strang is finding inverses at about 22:00?
(i just know the 18.02 method adj(a)/det(a) )

http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/factorization-into-a-lu/

Answer 27

is that the advantage of A=Lu ?

if we know the each of the elimination matrices E_32 , E_31, E_21
then L is just those three sort of superimposed with signs below the diagonal swapped?

Answer 28

@phi @ganeshie8