How do you find the #LU# factorization for #tildeA = [(1,0),(-3,1)]# such that #tildeL# is unit diagonal?

I had gotten an #LU# factorization, except my #tildeL# matrix wasn't lower-triangular, so it doesn't count. Here's the expression I used:

If #tildeA = tildeLtildeU#, then when #tildeE_ncdotcdotcdottildeE_2tildeE_1tildeA -> tildeU#:

#color(blue)(tildeA = stackrel(tildeL)(overbrace(tildeE_1^(-1)tildeE_2^(-1)cdotcdotcdottildeE_n^(-1)))tildeU)#.

where #tildeE# is an elementary matrix and #tildeE^(-1)# is its inverse. #tildeU# is an upper-triangular matrix, and #tildeL# is a lower-triangular matrix.

1 Answer
Jan 30, 2016

OH, I figured out what was going on. It turns out that the upper triangular matrix #tildeU# is allowed to be the identity matrix #[(1,0),(0,1)]#.

So, I already have #tildeL# apparently. Since #tildeA = tildeLtildeU# and #tildeU = tildeI#, #tildeA = tildeL#. Therefore:

#tildeL = [(1,0),(-3,1)]#
#tildeU = [(1,0),(0,1)]#

#tildeA = tildeLtildeU = [(1,0),(-3,1)][(1,0),(0,1)] = color(blue)([(1,0),(-3,1)])#


The longer way involves elementary matrices. Since each elementary row operation corresponds to the multiplication by an elementary matrix, we can do the following:

#stackrel(tildeA)(overbrace([(1,0),(-3,1)]))stackrel(3R_1 + R_2" ")(->)[(1,0),(0,1)] = tildeU#

Thus, #stackrel(tildeI)(overbrace([(1,0),(0,1)]))stackrel(3R_1 + R_2" ")(->) [(1,3),(0,1)] = tildeE_1#

So what we have is:

#tildeE_1tildeA = tildeU#

the condition for which #tildeA = tildeE_1^(-1)tildeE_2^(-1)cdotcdotcdottildeE_n^(-1)tildeU#.

As a result, we should realize that if we undo #tildeE_1# and transform it into the identity matrix #tildeI# using the inverse, #tildeE_1^(-1)#, we have the #LU# factorization.

#tildeE_1^(-1) = [(1,-3),(0,1)]#

#tildeE_1^(-1)tildeE_1tildeA = tildeE_1^(-1)tildeU#

#tildeItildeA = tildeA = tildeLtildeU = stackrel(tildeE_1^(-1))(overbrace([(1,0),(-3,1)]))stackrel(tildeU)(overbrace([(1,0),(0,1)])) = color(blue)(stackrel(tildeA)(overbrace([(1,0),(-3,1)])))#

so #tildeA = color(blue)(tildeL = [(1,0),(-3,1)])#.