In this problem, we will construct the QR factorization of the matrix A = [2 -4

Question

In this problem, we will construct the QR factorization of the matrix A = [2 -4 0 -2 6 10 1 2 2 0 -4 -2]. (a) Using the Gram-Schmidt process, find an orthogonal basis for Col(A). (b) Using your result from (a), find an orthonormal basis for Col(A). (c) Using your result from part (b), construct a matrix Q elementof R^4 times 3 with orthonormal columns such that Col(A) = Col(Q). (d) Verify that your matrix Q from part (c) satisfies the property that Q^T Q = I elementof R^3 times 3. (e) Using your matrix Q from part (c), find an upper triangular matrix R elementof R^3 times 3 such that A = QR.

Explanation / Answer

a)

v1= { 2 , -2 , 1, 0} v2={-4,6 2,-4} v={0 10 2 -2}

let u1= { 2 -2 1 0 }

u2= v2- proj < u1,v2 > = v2- < u1,v2> /< u1,u1> *u1

= { -4 , 6 ,2 ,-4 } - -18/9 { 2 ,-2 , 1 0 }= { 0,2,4,-4}

u3= v3-proj ( u1,v3 )-proj(u2,v3} = {0 10 2 -2} - -18 /9 { 2 ,-2 , 1 0 } - 36/36{ 0,2,4,-4} = { 4,, 4 , 0 , 2 }

thus orthogonal basis of col(A) is { u1 , u2 ,u3 }

b)

orthonormal basis of col(A) is { e1 ,e2, e3 }

where e= u/|| u||

thus e1= 1/3 { 2 -2 1 0 } e2= 1/6 { 0,2,4,-4} =1/3 { 0 , 1 ,2 ,-2 } e3= 1/6 { 4,, 4 , 0 , 2 } =1/3 { 2 , 2 ,0 , 1}

now

c)

Q={ e1; e2 ; e3 }

Q =1/3 [ 2 0 2 ; -2 1 2 ;1 2 0 ; 0 -2 1 }

d)

now QTQ= 1/3 [  2 -2 1 0 ] *1/3 [2 0 2 ] = 1/9 [9 0 0 ; 0 9 0 ; 0 0 9 ] = [ 1 0 0 ; 0 1 0 ; 0 0 1}= I

[0 , 1 ,2 ,-2 ] [-2 1 2 ]

[  2 , 2 ,0 , 1] [1 2 0 ]

[ 0 -2 1 ]

thus QTQ=I

Q-1=QT

e )

A=QR

R= Q-1A = QTA

R= 1/3 [  2 -2 1 0 ] * [ 2 -4 0 ] = [ 3 -6 -6 ]

[0 , 1 ,2 ,-2 ] [-2 6 10 ] [ 0 6 6 ]

[  2 , 2 ,0 , 1] [1 2 2 ] [ 0 0 6 ]

[0 -4 - 2 ]

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