Sorry about formatting as this is my first post here.

Question :- A mXn matrix (m>=n), A=Q'R' (Reduced QR)

To show that A has rank n iff diagonal entries of R' are non zero.

My Solution :-
if Ax = 0 => Q'R'x = 0 => R'x =0

let R' = upper triangular matrix (nXn) and x is a column vector.

suppose r_{nn} ..... r_{k+1k+1} are not 0

but r_{kk} = 0
since R'x = 0 => x_n = x_{n-1}..... = x_{k+1} = 0

but x_k is not zero as r_{kk} is 0 and we can have x_k= say 1 ....so we can determine value of x_{k-1} .... x_1 by back substitution


=> x is a nonzero vector such that R'x = 0


I am not able to go beyond this.....Am I taking right approach..

Any help is appreciated.