00:01
All right, problem 34, we have a matrix that is m by n and has rank positive and the action form of a.
00:24
So the first thing we need to do is to find an invertible matrix such that a equals e times u.
00:31
So recall, because u is the is the each in form of a, recall how we do the row reduction, and that is the same as we times a by some matrix e1 to get a new matrix, say u1, because the row reduction is a linear transformation and the rule reduction is clearly invertible.
01:09
So each of this e1 elementary matrix will be invertible.
01:17
And actually, usually the only once of the rule reduction will not be enough.
01:24
So we have plenty of a bunch of road reductions.
01:27
So that gives us a bunch of e1.
01:30
So that is, so i'll write it here.
01:36
So that means we have, say we have p times, p times.
01:43
We have p times rule reduction, so that's ep times until e1.
01:53
The operation of a gives u1, sorry, just u.
02:00
And since each of the elementary matrix is invertible, so the whole matrix is invertible.
02:08
So that means a is epp till e1 inverse times u.
02:19
So we just need to take e to be e p to e1 inverse.
02:34
So we're done.
02:36
Now to decompose a into several sum of rank 1 matrices, we need to apply the theorem 10 in our section 2 .4.
02:48
So the way to do that is to notice, well first we need, first we have a is equal to e times u.
02:59
So we write down the matrix e in the form of column vectors.
03:07
So we have e1...