# Why would the rref(A) * x still be 0 when A * x is 0?

1. Nov 2, 2005

### jtm

Assume x = rand(1)*v1 + rand(1)*v2 + rand(1)*v3 where each v is obtained by the columns of the nulbasis(A) which are the solutions to Ax = 0, and that rand(1) is a random coefficient (scalar). Why would the rref(A) * x still be 0 when A * x is 0? Thank you!

2. Nov 2, 2005

### Hurkyl

Staff Emeritus
Algebraically speaking, what is rref(A)?

3. Nov 2, 2005

### jtm

rref = reduced row echelon form

A =

3 2 9 6 7 9
6 5 18 15 6 6
4 9 12 27 8 2

R =

1.0000 0 3.0000 0 0 0.7432
0 1.0000 0 3.0000 0 -1.2973
0 0 0 0 1.0000 1.3378

Where R = rref(A)

4. Nov 2, 2005

### AKG

Unless I'm mistaken about what the row-reduced echelon form is precisely, rref(A) = UA for some invertible matrix U. U is a composition of elementary row operations (you are probably aware that you get to rref(A) by performing row operations on A... well these row operations can be represented by matrices). However, these extra details are irrelevant because if rref(A) = UA, then you don't need to know what U is like, since:

rref(A)(x) = (UA)(x) = U(A(x)) = U(0) = 0

5. Nov 3, 2005

### HallsofIvy

Yes. "Row reduction" can be thought of as applying a sequence of "row operations" to the matrix A but each row operation is equivalent to multiplying A by an "elementary matrix" (a matrix derived from the identity matrix by applying that row operation to it). That is, if there is a sequence of row operations, r1, r2, ..., rn, that reduces A to rref(A) (that's TI-85 notation isn't it?) then there exists a sequence of elementary matrices, m1, m2,...,mn such that the multiplication mn...m2m1A= rref(A). Obviously, if Ax= 0, then all the rest of the multiplications will give 0 also. Further, since no elementary matrix has determinant 0 (the inverse of an elementary matrix is the matrix corresponding to the "opposite" row operation) (other than "multiplying an entire row by 0" which most texts either never mention or don't consider a "row operation"), if rref(A)x= 0, then A= 0.