Question

Let {q1,q2,...,qn} be the orthonormal vectors obtain after applying the Gram-Schmidt procedure on an mxn matrix A. Show that A = QR, where Q is an mxn with columns q1,q2,...,qn and R is an nxn upper triangular matrix.

Answer 1

@amistre64

Answer 2

QR decomposition eh, i wouldnt be able to do this on the spot. I would have to read up on it to fill in the parts ive forgotten over theyears :)

Answer 3

yes..QR decomposition.. its okay..thank you :)

Answer 4

Q = [e1, e2, e3, ..., en]

R = [a1.e1; a2.e1; a3.e1; ...; an.e1 ]
[ 0 ; a2.e2; a3.e2; ...; an.e2 ]
[ 0 ; 0 ; a3.e3; ...; an.e2 ]
[ ........................................... ]
[ 0 ; 0 ; 0 ; ...; an.en ]

Answer 5

Show that: ak = [e1; e2; e3;...; ek; ....;en] * [ak.e1; ak.e2; ak.e3; ...; ak.ek; 0; 0; ....; 0]

Answer 6

well, since everything after ak.ek is 0; we can effectively ignore it. So demonstrate that:

ak = [e1; e2; e3;...; ek] * [ak.e1; ak.e2; ak.e3; ...; ak.ek]

Answer 7

the dot product of ak.er is equal to the rth element of ak

giving us a vector of say: [ak1; ak2; ak3; ak4; ...; akk] does that make sense?

Answer 8

my Q row is off,

row1 is equal to e1: 1,0,0,0,...,0
row2 is equal to e2: 0,1,0,0,...,0
etc

Answer 9

ak = [ek] * [ak.e1; ak.e2; ak.e3; ...; ak.ek]

ak = [ek] * [ak1; ak2; ak3; ...; akk]

ak = the kth element of ak

but this might be way too much work to show and they might be expecting you to use thrms that i have long forgotten

Answer 10

okay..thanks a lot :)

Answer 11

@SolomonZelman

Answer 12

@phi

Answer 13

@radar

Answer 14

Sorry, this is above my pay-grade. Haven't a clue.

Answer 15

its okay..thank you @radar :)

Answer 16

This is a pain to explain.
call the columns of A= a1, a2, etc

The idea is to use the definition of the Gram-Schmidt procedure to show that q1 is a1 (normalized) , a2 = a1 - a2 projected onto a1,
a3 = a3 - a3 proj a1 - a3 proj a2

Set up the equation
A = Q R
the columns of R will select out the columns of Q to combine
a1 (first col of A) = q1 scaled by |a1|, and no other columns.
i.e. the first column of R will have 1 entry at 1,1, and zeros below it.

a2 will be a linear combination of q1 and q2 only...
if you follow this pattern you find R is upper triangular.

Answer 17

I knew this was above my paygrade.