4.1
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.