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MATH 415, SECTION Q
REVIEW SHEET 2
SOLUTIONS
BERT GUILLOU
True or False
- Let v 1 ,... , vn be nonzero vectors in an inner product space V such that each pair of vectors vi and vj is orthogonal. Then the vectors are linearly independent.
True We discussed this in class, and it is given as Proposition 5.4 in the text. The essential idea is that if you have a linear combination
0 = c 1 v 1 + · · · + cnvn,
then by considering the inner product of the above expression with vi, you can show that ci = 0.
- Let C be the positive-definite matrix corresponding to some inner product on Rn^ and let v 1 ,... , vk be vectors (in Rn). If A denotes the matrix with columns the vi’s, then the Gram matrix K = AT^ CA is positive-definite.
False Theorem 3.28 says that a Gram matrix is always positive semi-definite, but in order to guarantee that it is positive definite, the vectors v 1 ,... , vn must be linearly independent. Otherwise, the matrix A has a kernel, and any vector in ker A gives a null vector for the Gram matrix.
- Let V be a finite-dimensional inner product space. The only subspace W ⊆ V such that W ⊥^ = { 0 } is W = V.
True Suppose that W ⊆ V satisfies W ⊥^ = { 0 }. Then by the unique decomposition result (Proposition 5.49 in the text), any vector w can be written uniquely as v = w + z with w ∈ W and z ∈ W ⊥. Since W ⊥^ = { 0 }, z = 0 and v = w ∈ W.
Date: December 12, 2008.
2 BERT GUILLOU
Problems
- Which of the following determine inner products on R^3? (i) x 1 y 1 − x 2 y 2 + 3x 3 y 3
(ii) x 1 y 1 + 2x 2 y 2 + 3x 2 y 3 + 12 x 3 y 3
(iii) x 1 y 1 + 2x 1 y 2 + 2x 2 y 1 + 3x 2 y 2 + x 2 y 3 + x 3 y 2 + 2x 3 y 3
(iv) x 1 y 1 + x 1 y 2 + x 2 y 1 + 2x 2 y 2 − x 2 y 3 − x 3 y 2 + 3x 3 y 3
Solution. (i) is not an inner product since it does not satisfy positivity: 〈
(ii) is not an inner product since it is not symmetric (there is a x 2 y 3 term but no x 3 y 2 term). For example, 〈
(iii) is not an inner product since it does not satisfy positivity. To see this, it is simplest to rewrite this as the matrix
C =
and check whether the matrix is regular with positive pivots. We find
so positivity fails. Indeed, 〈
(iv) is an inner product. Bilinearity and symmetry are simple to check, and we check positivity by forming the appropriate matrix as above. We have
C =
so positivity holds.
4 BERT GUILLOU
(i)
1 2 −^2
from which we find that
2 x^2 + 8xy + 3y^2 + 2xz + 24yz + z^2 = 2(x + 2y +
z)^2 − 5(y − 2 z)^2 +
z^2.
(ii)
from which we find that
x^2 + 6xy + 5y^2 + 8yz + z^2 + 4xw − 4 yw + 6zw + 7w^2 = (x + 3y + 2w)^2 − 4(y − z + 2w)^2 + 5(z − w)^2 + 14w^2. (iii)
3 2 1 0 1 34 1
from which we find that
2 x^2 + 6xy +
y^2 + 4xz + 12yz +
z^2 = 2
x +
y + z
y +
z
(iv)
from which we find that
x^2 + 8xy + 13y^2 − 4 xz − 4 yz − 8 z^2 = (x + 4y − 2 z)^2 − 3 (y − 2 z)^2.
- Give the Gram matrix associated to the following sets of vectors and choices of inner products
(i) the vectors
, using the weighted inner product 3x 1 y 1 +x 2 y 2 +4x 3 y 3
MATH 415, SECTION Q1 REVIEW SHEET 2 SOLUTIONS 5
(ii) the vectors
, using the inner product given by C =
(iii) the vectors
and
, using the dot product
(iv) the vectors 1, 1 + x, and 1 − x^2 in the vector space P of polynomials on the interval [0, 2], using the L^2 inner product defined by 〈p(x), q(x)〉 =
0 p(x)q(x)^ dx.
Solution. The Gram matrix can be computed either by the explicit formula Kij = 〈vi, vj 〉 or by the formula K = AT^ CA. (i) We compute 〈v 1 , v 1 〉 = 4, 〈v 1 , v 2 〉 = 2, 〈v 1 , v 3 〉 = −3, 〈v 2 , v 2 〉 = 40, 〈v 2 , v 3 〉 = −20, and 〈v 3 , v 3 〉 = 19. (The other inner products are given by symmetry.) The Gram matrix is
K =
(ii) The Gram matrix is given by
K =
T C
(iii) We compute 〈v 1 , v 1 〉 = 6, 〈v 1 , v 2 〉 = 7, and 〈v 2 , v 2 〉 = 15. The Gram matrix is then
K =
(iv) We compute 〈 1 , 1 〉 = 2, 〈 1 , 1 + x〉 = 4, 〈 1 , 1 − x^2 〉 = −^23 , 〈1 + x, 1 + x〉 = 263 , 〈1 + x, 1 − x^2 〉 = −^83 , and 〈 1 − x^2 , 1 − x^2 〉 = 4615. We conclude that the Gram matrix is
K =
2 4 −^23
4 263 −^83
−^23 −^834615
- Determine which of the following quadratic functions have minimum values. For those that do attain a minimum, say where the function hits the minimum (the minimizer) and give the corresponding minimum value.
(i) 2x^2 + 2xy + y^2 − y + 3
MATH 415, SECTION Q1 REVIEW SHEET 2 SOLUTIONS 7
(i) A =
(^) and b =
, so AT^ A =
and AT^ b =
. We then reduce
the augmented matrix: ( 58 17 14 17 6 5
This gives w 2 = 4247 and w 1 = − 471.
(ii) A =
,^ b^ =
, so^ A
T A =
and AT^ b =
. We then reduce
the augmented matrix: (^) (
25 3 16 3 38 2
Then w 2 = 9412 and w 1 = 1423 ,,^956525.
(iii) AT^ A =
(^) and AT^ b =
. We reduce the augmented matrix
0 343 − 3 −^23
Then w 3 = 164959 , w 2 = − 95913 , and w 1 = 324959.
- Use the Gram-Schmidt process to convert the sets of vectors given in problem 4 into orthogonal sets of vectors, under the given inner products.
Solution. (i) Let w 1 , w 2 , and w 3 denote the original vectors. As usual, we take v 1 = w 1 and find ‖v 1 ‖^2 = 3 · 12 + 1^2 + 4 · 02 = 4. We then have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
=^1
and
‖v 2 ‖^2 =
(3 · (−1)^2 + 3^2 + 4 · 62 ) = 39.
Finally,
v 3 = w 3 −
〈v 1 , w 3 〉 ‖v 1 ‖^2
v 1 −
〈v 2 , w 3 〉 ‖v 2 ‖^2
v 2 =
− −^3
1 2 (−45) 39
8 BERT GUILLOU
(ii) Let w 1 , w 2 , and w 3 denote the original vectors. As usual, we take v 1 = w 1 and find ‖v 1 ‖^2 = 1 · 02 + 2 · 0 · 1 + 2 · 12 + 1 · 22 = 6. We then have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
=^1
and
‖v 2 ‖^2 =
(6^2 + 2(6)(−4) + 2 · (−4)^2 + 1^2 ) =
Finally,
v 3 = w 3 −
〈v 1 , w 3 〉 ‖v 1 ‖^2
v 1 −
〈v 2 , w 3 〉 ‖v 2 ‖^2
v 2 =
1 3 ·^4 7 3
=^10
(iii) Let w 1 and w 2 denote the original vectors. As usual, we take v 1 = w 1 and find ‖v 1 ‖^2 = 1^2 + 0^2 + (−1)^2 + 2^2 + 0^2 = 6. We then have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
(iv) Let w 1 , w 2 , and w 3 denote the original vectors. As usual, we take v 1 = w 1 and find ‖v 1 ‖^2 =
0 1 dx^ = 2. We then have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 = 1 + x −
1 = x − 1
and
‖v 2 ‖^2 =
0
(x − 1)^2 dx =
Finally,
v 3 = w 3 −
〈v 1 , w 3 〉 ‖v 1 ‖^2
v 1 −
〈v 2 , w 3 〉 ‖v 2 ‖^2
v 2 = 1 − x^2 −
−^23
−^43
2 3
· (x − 1) = −x^2 + 2x −
- Find an orthonormal basis for R^3 under the inner product given by C =
and also under the inner product x 1 y 1 +−x 1 y 3 −x 3 y 1 +2x 2 y 2 +2x 2 y 3 +2x 3 y 2 +4x 3 y 3. Express
the vectors
(^) and
(^) in terms of these bases.
10 BERT GUILLOU
It follows that
(^) = 2v 1 + 2v 2 + 2v 3. Similarly, we find
〈w, v 1 〉 ‖v 1 ‖^2
〈w, v 2 〉 ‖v 2 ‖^2
and 〈w, v 3 〉 ‖v 3 ‖^2
so that
(^) = 9v 1 + 6v 2 + v 3.
Using now the second inner product, we first compute ‖v 3 ‖^2 = 1. We then have 〈u, v 1 〉 ‖v 1 ‖^2
〈u, v 2 〉 ‖v 2 ‖^2
and 〈u, v 3 〉 ‖v 3 ‖^2
It follows that
(^) = − 2 v 1 + 4v 2 + 2v 3. Similarly, we find
〈w, v 1 〉 ‖v 1 ‖^2
〈w, v 2 〉 ‖v 2 ‖^2
and 〈w, v 3 〉 ‖v 3 ‖^2
so that
(^) = 2v 1 + 7v 2 + v 3.
- Find the closest point in the subspace W to the point β.
(i) W = span
⊆ R^3 and β =
(ii) W = span
⊆ R^4 and β =
MATH 415, SECTION Q1 REVIEW SHEET 2 SOLUTIONS 11
(iii) redo (i) using the weighted inner product 2x 1 y 1 + 3x 2 y 2 + 5x 3 y 3
(iv) redo (ii) using the weighted inner product 2x 1 y 1 + 12 x 2 y 2 + 4x 3 y 3 + 3x 4 y 4
Solution. We discussed two methods of finding the closest point. The first, in Chapter 4, was in the context of minimization problems. The second method, from Chapter 5, was through the technique of orthogonal projection. In fact, we saw that orthogonal projection was just another name for the closest point. To illustrate the two methods, we will use the first method for (i) and (iii) and the second for (ii) and (iv). (i) The method from Chapter 4 says that the closest point can be found as follows. Let
w 1 =
(^) and w 2 =
. Then we are looking for a vector w = c 1 w 1 + c 2 w 2. The
coefficient vector c is given as the solution to the matrix equation AT^ Ac = AT^ β, where A is the matrix with columns w 1 and w 2. Here we have
AT^ A =
and AT^ β =
We then reduce the augmented matrix ( 5 2 1 2 10 1
We conclude that c 2 = 463 and c 1 = 11520. So the closest point to
(^) is
w 1 +
w 2 =
(ii) The method from Chapter 5 says to first convert the basis to an orthogonal basis via the Gram-Schmidt process. We first set v 1 = w 1 and compute ‖v 1 ‖^2 = 11. We then have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
−^
=^
and
‖v 2 ‖^2 =
[
(−13)^2 + (35)^2 + (−22)^2 + (16)^2
]
MATH 415, SECTION Q1 REVIEW SHEET 2 SOLUTIONS 13
(iv) Again, we first set v 1 = w 1 and compute ‖v 1 ‖^2 = 592. We then have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
−^
59 2
=^
and
‖v 2 ‖^2 =
[
2(−88)^2 +
(206)^2 + 4(−118)^2 + 3(31)^2
]
Finally,
v 3 = w 3 −
〈v 1 , w 3 〉 ‖v 1 ‖^2
v 1 −
〈v 2 , w 3 〉 ‖v 2 ‖^2
v 2 =
17 2 59 2
1 59 ·^196 1615 59
=^
and
‖v 3 ‖^2 =
We then compute the appropriate inner products with β:
c 1 =
〈v 1 , β〉 ‖v 1 ‖^2
59 2
, c 2 =
〈v 2 , β〉 ‖v 2 ‖^2
− 814 59 1615 59
c 3 =
〈v 3 , β〉 ‖v 3 ‖^2
1226 1615 1046 1615
So the closest point to
under the weighted inner product is
v 1 +
v 2 +
v 3 =
=^
- Find the orthogonal projection of the vector v onto the subspace W when
(i) W = span
⊆ R^3 and v =
14 BERT GUILLOU
(ii) W = span
⊆ R^4 and v =
(iii) W = span
⊆ R^4 and v =
Solution. The method is the same as that employed in parts (ii) and (iv) of Problem 9 above. (i) We first find an orthogonal basis. Let v 1 = w 1. Then ‖v 1 ‖^2 = 2. We have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
=^1
and
‖v 2 ‖^2 =
[
12 + (−1)^2
]
We then compute the appropriate inner products with v:
c 1 =
〈v 1 , v〉 ‖v 1 ‖^2
= 3, c 2 =
〈v 2 , v〉 ‖v 2 ‖^2
1 2 ·^4 1 2
So the closest point to
(^) in W is
3 v 1 + 4v 2 =
(ii) Let v 1 = w 1. Then ‖v 1 ‖^2 = 31. We have
v 2 = w 2 −
〈v 1 , w 2 〉 ‖v 1 ‖^2
v 1 =
−^
=^
and
‖v 2 ‖^2 =
[
(24)^2 + (69)^2 + (−45)^2 + (27)^2
]
16 BERT GUILLOU
So the closest point to
in^ W^ is
v 1 +
v 2 +
v 3 =
=^
