16. Let v_4 =

w x y z Then the dot product of v_4 with each of the three given vectors v_1, v_2, v_3 must be 0, which leads to the system of linear equations:

(1/2)w + (1/2)x + (1/2)y + (1/2)z = 0 (1/2)w + (1/2)x - (1/2)y - (1/2)z = 0 (1/2)w - (1/2)x + (1/2)y - (1/2)z = 0

which has matrix

1/2 1/2 1/2 1/2 0 1/2 1/2 -1/2 -1/2 0 1/2 -1/2 1/2 -1/2 0

We find the RREF. Doubling the first row and the subtracting half of it from the second and third rows yields:

1 1 1 1 0 0 0 -1 -1 0 0 -1 0 -1 0

Interchanging the second and third rows and multipling the second and third rows by -1 produces:

1 1 1 1 0 0 1 0 1 0 0 0 1 1 0

Subtracting the second row from the first gives

1 0 1 0 0 0 1 0 1 0 0 0 1 1 0

Finally, subtracting the third row from the first gives

1 0 0 -1 0 0 1 0 1 0 0 0 1 1 0

which is the RREF. The general solution is

z { -z : z in R } -z z

which the same as the span of the vector

1 -1 -1 1

But we need a vector of length one, and this vector has length (1 + 1 + 1 + 1)^(1/2) = 4^(1/2) = 2. There are two solutions: the vector

1/2 -1/2 -1/2 1/2

and its negative

-1/2 1/2 1/2 -1/2

28. We get an orthonormal basis for the subspace spanned by the three given vectors if we take half of each of them, i.e., the three vectors u, v, w given by

1/2 1/2 1/2 1/2 , 1/2 , and -1/2 1/2 -1/2 -1/2 1/2 -1/2 1/2

respectively are mutually perpendicular and of length one. Thus, the orthogonal projection of z =

1 0 0 0

on their span is given by

(z.u)u + (z.w)v + (z.w) = (1/2)u + (1/2)v + (1/2)w = (1/2)(u + v + w), and since u + v + w =

3/2 1/2 -1/2 1/2

the projection is

3/4 1/4 -1/4 1/4

4.2

4. + 18. We perform the Gram-Schmidt process and find the
QR factorization simultaneously. Call the vectors v_1, v_2, v_3.
v_1 has length 5, and so w_1 = (1/5)v_1 =

4/5 0 3/5

and this also tells us that r_{11} = 5.

Then (v_2 . w_1) = 20 + 0 - 15 = 5, which tells us that r_{12} = 5, and to find w_2 we first find v_2 - 5 w_1 =

21 0 -28

and then divide by its length. Since this vector is 7 times the vector

3 0 -4

its length is 35. This tells us that r_{22} = 35 and that w_2 =

21/35 3/5 0 = 0 -28/35 -4/5

Then r_{13} = (v_3 . w_1) = 0, r_23 = (v_3 . w_2) = 0, and so to find w_3 we simply divide v_3 - 0 - 0 = v_3 by its length, which is r_{33} = 2, and we have that w_3 =

0 -1 0

We have now completed both tasks. Explicitly, the QR factorization is

4 25 0 4/5 3/5 0 5 5 0 0 0 -2 = 0 0 -1 0 35 0 3 -25 0 3/5 -4/5 0 0 0 2

8. + 22. Call the vectors v_1 and v_2. The length of v_1 is (25 + 16 + 4 + 4)^(1/2) = 49^(1/2) = 7, and so r_{11} = 7 and w_1 =

5/7 4/7 2/7 2/7

Then r_{12} = (v_2 . w_1) = 15/7 + 24/7 + 14/7 - 4/7 = 49/7 = 7, and to find w_2 we first find v_2 - 7w_1 =

3 5 -2 6 - 4 = 2 7 2 5 -2 2 -4

We now divide this vector by its length, which is 7, and which is also r_{22}, to get w_2 =

-2/7 2/7 5/7 -4/7

and we are now done. Explicitly, the QR factorization is

5 3 5/7 -2/7 7 7 4 6 = 4/7 2/7 0 7 2 7 2/7 5/7 2 -2 2/7 -4/7

34. We get the RREF of the given matrix by subtacting the first row from the second to get

1 1 1 1 0 1 2 3

and then the second row from the first:

1 0 -1 -2 0 1 2 3

and so the kernel, which is the general solution set of the corresponding homogeeous system of linear equations (calling the variables w, x, y, z) is

y + 2z { -2y - 3z : y, z in R } y z

which is the span of the vectors

1 2 -2 and -3 1 0 0 1

Call these v_1 and v_2.

The length of the first vector is (1 + 4 + 1)^{1/2} = 6^(1/2). Thus, we may take w_1 =

1/6^(1/2) -2/6^(1/2) 1/6^(1/2) 0

Then (v_2 . w_1) = (2 + 6 + 0 + 0)/6^(1/2) = 8/6^(1/2), and v_2 - (8/6^(1/2))w_1 =

2 8/6 2/3 -3 - -16/6 = -1/3 0 8/6 -4/3 1 0 1

The length of this vector is (4/9 + 1/9 + 16/9 + 1)^(1/2) = (30/9)^(1/2) =(30)^(1/2)/3, and dividing we get that w_2 =

2/30^(1/2) -1/30^(1/2) -4/30^(1/2) 3/30^(1/2)

w_1, w_2 is the required orthonormal basis for the kernel.

4.3

12. The vector

a

b

c

must be orthogonal to the other two columns. Taking the dot products and clearing denominators (multiply the first equation by 3 and the second by 2^(1/2) ) we get:

2a + 2b + c = 0 a - b = 0

Putting the matrix

2 2 1 0 1 -1 0 0

in RREF (divide the first row by 2 and subtract from the second) we get

1 1 1/2 0 0 -2 -1/2 0

Divide the second row by -2 and subtract from the first to get:

1 0 1/4 0 0 1 1/4 0

and so the general solution is

-c/4 { -c/4 : c in R } c

which is the span of

-1/4 -1/4 1

The possible solutions will be the multiples of this vector of length one. The length of the vector is (1/16 + 1/16 + 1)^(1/2) = (1/8 + 1)^(1/2) = (9/8)^(1/2) = 3/2(2^(1/2)).

Dividing by this length, we find that one possibility for the third column is

-2^(1/2)/6 -2^(1/2)/6 2(2^(1/2))/3

The only other possibility is the negative of this vector.

26. Say Q is n by m. Since Q 's columns are an orthonormal set, Q^T Q = 1_m. Multiplying both sides of M = QR by Q^T on the left we get that Q^T M = Q^T (QR) = (Q^T Q)R = (1_m)R = R, as required.