This one involves three vectors. Let v_1 =

3

4

0

0

v_2 =

3

4

6

8

and v_3 =

2

1

2

1

We want to find an orthonormal basis for the span V of v_1, v_2, v_3, and we also want the QR factorization of the matrix formed from the v's, namely

3 3 2

4 4 1

0 6 2

0 8 1

Note that Q will be the 4 by 3 matrix formed from the w's, while R = [r_ij] will be a 3 by 3 upper triangular matrix whose entries are computed in the course of finding the w's.

w_1 = (1/|| v_1 ||) v_1 = (1/5)v_1 =

3/5

4/5

0

0

and this also tells us that $r_11 = || v_1 || = 5.

To find w_2 first subtract the projection of v_2 on the line through w_1. Since v_2 . w_1 = 25/5 = 5, which is also r_12, the projection is 5w_1 which is

3

4

0

0

and subtracting this from v_2 gives

0

0

6

8

The length of this vector is r_22 = 10 and dividing by it gives w_2 which is

0

0

3/5

4/5

The dot product of v_3 and w_1 is 6/5 + 4/5 = 2 = r_13,

the dot product of v_3 and w_2 is 6/5 + 4/5 = 2 = r_23,

and then v_3 - 2 w_1 - 2 w_2 =

2 6/5 0

1 - 8/5 - 0

2 0 6/5

1 0 8/5

which is

4/5

-3/5

4/5

-3/5

The length of this vector is the square root of (16+9+16+9)/25 , i.e., the square root of 2, and so r_33 = 2^.5. Dividing the vector above by 2^.5, which is the same as multiplying by 2^.5/2, we find that w_3 =

4(2^.5)/10

-3(2^.5)/10

4(2^.5)/10

-3(2^.5)/10

We could make use of the fact that 4(2^.5)/10 = 2(2^.5)/5 but we shall leave w_3 as is. Thus, Q is the matrix with columns w_1, w_2, w_3 and is

3/5 0 4(2^.5)/10

4/5 0 -3(2^.5)/10

0 3/5 4(2^.5)/10

0 4/5 -3(2^.5)/10

and R is the matrix (entries computed as we went along):

r_11 r_12 r_13

0 r_22 r_23

0 0 r_33

or

5 5 2

0 10 2

0 0 2^.5

One can check that QR =

3 3 2

4 4 1

0 6 2

0 8 1

as required.

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