11.10. EXERCISES 291
Proof: It is of the form∑n
i=1 viCi where Ci is a suitable (n− 1)× (n− 1) determinant.Thus the inner product of this with vk for k ≤ n − 1 is the expansion of a determinantwhich has two equal columns. However, the inner product with vn will be the Grammianof {v1, ..., vn} which is not zero since these vectors vi are independent. See Problem 11 onPage 293. ■
Example 11.9.2 The vectors 1, x, x2, x3 are linearly independent on [0, 1], the vector spacebeing the continuous functions defined on [0, 1]. You might show this. An inner product is
given by∫ 1
0f (x) g (x) dx. Find an orthogonal basis for span
(1, x, x2, x3
).
You could use the above lemma. u1 (x) = 1. Now I will assemble the formal determinantsas given above.
det
(1 112 x
),det
1 12 1
12
13 x
13
14 x2
, det
1 1
213 1
12
13
14 x
13
14
15 x2
14
15
16 x3
Now the orthogonal basis is obtained from evaluating these determinants and adding 1
to the list. Thus an orthonormal basis is{1, x− 1
2 ,112x
2 − 112x+ 1
72 ,1
2160x3 − 1
1440x2 + 1
3600x− 143 200
}Is this horrible? Yes it is. However, if you have a computer algebra system do it for you,
it isn’t so bad. For example, to get the last term, you just do1
x
x2
x3
( 1 x x2)=
1 x x2
x x2 x3
x2 x3 x4
x3 x4 x5
Then you do the following.
∫ 1
0
1 x x2
x x2 x3
x2 x3 x4
x3 x4 x5
dx =
1 1
213
12
13
14
13
14
15
14
15
16
You could get Matlab to do it for you. Then you add in the last column which consists ofthe original vectors. If you wanted an orthonormal basis, you could divide each vector byits magnitude. This was only painless because I let the computer do all the tedious busywork. However, I think it has independent interest because it gives a formula for a vectorwhich will be orthogonal to a given set of linearly independent vectors.
11.10 Exercises
1. Here are three vectors in R4 : (1, 2, 0, 3)T, (2, 1,−3, 2)
T, (0, 0, 1, 2)
T. Find the three
dimensional volume of the parallelepiped determined by these three vectors.
2. Here are two vectors in R4 : (1, 2, 0, 3)T, (2, 1,−3, 2)
T. Find the volume of the paral-
lelepiped determined by these two vectors.
3. Here are three vectors in R2 : (1, 2)T, (2, 1)
T, (0, 1)
T. Find the three dimensional
volume of the parallelepiped determined by these three vectors. Recall that from theabove theorem, this should equal 0.