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MATH 415, SECTION D
REVIEW SHEET 2
SOLUTIONS
BERT GUILLOU
True or False
- Let v 1 ,... , vn be nonzero vectors in Rn^ such that each pair of vectors vi and vj is orthogonal. Then the vectors are linearly independent.
True. Suppose that c 1 v 1 + · · · + cnvn = 0.
Now take the dot product of both sides with the vector v 1. The left-hand side gives c 1 ‖v 1 ‖^2 , whereas the right-hand side certainly gives 0. So c 1 = 0. Taking dot products with some other vi shows that ci = 0. Since all ci’s must be 0, the vectors are linearly independent.
- The only subspace V ⊆ Rn^ such that V ⊥^ = { 0 } is V = Rn.
True. Suppose that x is a vector in Rn. Then x can be written as x = v + e,
where v is in V and e is in V ⊥. But then e must be 0 , so x = v is in V. Since we have shown that every x in Rn^ is in V , it follows that V is Rn.
- Let A be an n × m matrix. If dim N (A) = k, then dim N (AT^ ) = n − m + k.
True. This follows from the Fundamental Theorem. The Theorem says that dim N (A) = m − r, where r is the rank. Since dim N (A) = k, it follows that r = m − k. But the Fundamental Theorem also says that dim N (AT^ ) = n − r, so we find that dim N (AT^ ) = n − m + k.
- If {v 1 ,... , vk} is an orthonormal basis for a subspace V of Rn^ and {w 1 ,... , wn−k} is an orthonormal basis for the complement V ⊥, then {v 1 ,... , vk, w 1 ,... , wn−k} is an orthonormal basis for Rn.
True. We essentially showed this in class. Since the set {v 1 ,... , vn, w 1 ,... , wn−k} con- sists of n vectors, it suffices to show the vectors are linearly independent. We already know the v’s are independent. But w 1 is orthogonal to all of the v’s, so by the argument in
Date: October 20, 2009.
2 BERT GUILLOU
problem 1 above, the vectors {v 1 ,... , vk, w 1 } are independent. The same argument works for adding in the vector w 2 as well as the rest of the w′s.
- If the columns of the n × n matrix A form an orthogonal basis for Rn, then the matrix A is an orthogonal matrix.
False. The terminology may be slightly confusing, but the definition of an orthogonal basis is that its columns form an orthonormal basis for Rn. Just orthogonal is not enough.
For instance,
is an orthogonal, but not orthonormal, basis for R^2. We can see
that the matrix A =
is not orthogonal since
AT^ A =
is not the identity matrix I 2.
- A reflection matrix is always an orthogonal matrix.
True. We discussed this in class. Since the reflection through a subspace V is given by R = 2P − I, where P is the projection matrix, we have
RT^ R = (2P T^ − IT^ )(2P − I) = (2P − I)(2P − I) = 4P 2 − 2 P − 2 P + I = 4P − 4 P + I = I.
This uses that every projection matrix is symmetric and satisfies P 2 = P.
- A projection matrix is always an orthogonal matrix.
False. A projectin matrix is not even invertible, as any vector orthogonal to the subspace V is projected to the 0 vector.
4 BERT GUILLOU
(i) (corrected) A =
(^) and b =
, so AT^ A =
and AT^ b =
We then reduce the augmented matrix: ( 58 17 14 17 14 7
This gives x 2 = 168523 and x 1 = 52377.
(ii) (corrected) A =
,^ b^ =
, so^ A
T A =
and AT^ b =
. We
then reduce the augmented matrix: ( 25 3 16 3 38 2
Then x 2 = 9412 and x 1 = 602941.
(iii) AT^ A =
(^) and AT^ b =
. We reduce the augmented matrix
0 343 − 3 −^23
Then x 3 = 164959 , x 2 = − 95913 , and x 1 = 324959.
- Find the least squares lines for the following sets of data points, and compute ‖e‖^2 , where e = Axˆ − b is the error vector.
(i) (0, 1), (1, 1), and (2, 3).
(ii) (0, 1), (1, 3), (2, 2), and (3, 3).
(iii) (0, 2), (2, 3), and (4, 4).
Solution. We want to find the line y = mx + c which comes closest to fitting the data points.
(i) Here A =
(^) and we are looking for the least squares solution to the equation
m c
MATH 415, SECTION D1 REVIEW SHEET 2 SOLUTIONS 5
The least squares solution is given by the solution to the normal equation
( 0 1 2 1 1 1
m c
or (^) ( 5 3 3 3
m c
Solving this gives
m c
. The least squares line is y = x + 2/3. The error vector is
e =
=^1
so that
‖e‖^2 =
(ii) Here A =
and we are looking for the least squares solution to the equation
m c
The least squares solution is given by the solution to the normal equation
( 0 1 2 3 1 1 1 1
m c
or (^) ( 14 6 6 4
m c
Solving this gives
m c
. The least squares line is y = 12 x + 32. The error vector is
e =
MATH 415, SECTION D1 REVIEW SHEET 2 SOLUTIONS 7
(iii) Any permutation matrix is an orthogonal matrix.
- Use the Gram-Schmidt process to convert the following bases for R^3 into orthonormal
bases. Also, write the vector e 1 =
(^) in terms of the new basis.
(i)
(ii)
Solution. (i) As a first step, we let w 1 =
(^) and find that ‖w 1 ‖^2 = 2. Next, set
w 2 =
Then ‖w 2 ‖^2 = 11. Finally,
w 3 =
− −^1
− −^5
The orthonormal basis is then
, √^1
, √^1
The expression for e 1 in terms of this basis is
e 1 =
(ii) As a first step, we let w 1 =
(^) and find that ‖w 1 ‖^2 = 14. Next, set
w 2 =
8 BERT GUILLOU
Then ‖w 2 ‖^2 = 3. Finally,
w 3 =
− −^1
=^11
The orthonormal basis is then
, √^1
, √^1
The expression for e 1 in terms of this basis is
e 1 =
+^1
+^1
- Find the matrix corresponding to projection onto the subspace with basis given by
(i)
(^) and
(ii)
and
Solution. The general formula is
P = A(AT^ A)−^1 AT^.
(i) Here
A =
so that
AT^ A =
The inverse is given by
(AT^ A)−^1 =
10 BERT GUILLOU
Solution. This is the projection of b onto W. The matrix A is
, and we find
that (AT^ A)−^1 = 461
. Then
P b =
