**Unformatted text preview: **Version 073 – EXAM04 – gilbert – (56780)
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001 10.0 points If B = P DP −1 with P an orthogonal matrix and D a diagonal matrix, then B is a
symmetric matrix. 1
8
2 −17 5. Explanation:
The normal equations for a least-squares
solution of Ax = b are by deﬁnition
AT Ax = AT b.
Now, True or False?
1. FALSE AT A = 2. TRUE correct
Explanation:
Note that for B to be symmetric, B T = B.
Also note that all diagonal matrices are by
deﬁnition symmetric. Consider B T .
B T = (P DP T )T = (P T )T DT P T
= P DT P T = P DP T = B.
Thus B T = B and B is by deﬁnition symmetric.
Consequently, the statement is
TRUE .
002 2. 1 3
correct
2 −8 6
7 2
−1 6
−2 −2
1 1
1 2
0
1 0
−1 0 and
AT b = 1
0 2
−1 6
17
1 .
7 =
−7
0
−3 Hence the least squares solution of Ax = b is
the solution x to the equation
6
−2 17
−2
.
x=
−7
1 This can be solved with row reduction or inverse matrices to determine that the solution
is
1 1 2
2 2 6
1 3
.
=
2 −8 (AT A)−1(AT b) = 17
−7 Consequently, the least squares solution to
Ax = b is 17
−7 −5
3.
9
4. = 1
0 10.0 points Find the least-squares solution of Ax = b
when 1 0
6
A = 2 −1 , b = 7 .
1 0
−3
1. 1 1 3
2 −8
003 . 10.0 points Determine the singular value σ1 for the matrix 2 0
A = 0 −2 .
1 2 Version 073 – EXAM04 – gilbert – (56780)
1. σ1 = 2 Since V has these three properties, V is a
subspace of itself. √ 7
√ Consequently, the statement is 2. σ1 = 2 2
√
3. σ1 = 5 TRUE . 4. σ1 = 3 correct
√
5. σ1 = 6 005
When Explanation:
By deﬁnition, σ is a singular value of A
√
when σ = λ and λ is an eigenvalue of AT A;
σ1 is the largest of these singular values.
Now
2 0
0 −2 AT A = 2 0
1 0 −2 2
1 2 5 2
.
2 8 = 5−λ
2 2
8−λ = λ2 − 13λ + 36 = (λ − 9)(λ − 4) .
Consequently,
σ1 = 3 .
004 1
,
−2 u1 = u2 = 2
,
1 are eigenvectors of a symmetric 2 × 2 matrix
A corresponding to eigenvalues
λ1 = 2 , λ2 = −3 , ﬁnd matrices D and P in an orthogonal diagonalization of A.
1. D = But then
det(AT A − λI) = 10.0 points 1 −2
2 0
, P =√
0 −3
5 1 2. D = −2
2 0
, P =
1
0 −3 1
2 3. D = 1
2 0
, P =
−2
0 −3 2
1 4. D = −3
0 1
0
, P =
−2
2 2
1 5. D = −3
0 1
0
1
, P =√
2
5 −2 10.0 points A vector space V is a subspace of itself.
True or False?
1. FALSE
2. TRUE correct
Explanation:
A set H is a subspace of a vector space V
when
(i) H contains the zero vector 0,
(ii) the sum u + v of any u, v in H is in H,
(iii) the scalar multiple cu of any scalar c
and any u in H is in H. 6. D = correct 2 0
,
0 −3 1
2 2
1 1
1
P = √
−2
5 2
1 Explanation:
When
D= λ1
0 0
, Q = [u1 u2 ],
λ then Q has orthogonal columns and A =
QDQ−1 is a diagonalization of A, but it is Version 073 – EXAM04 – gilbert – (56780)
not an orthogonal diagonalization because Q
is not an orthogonal matrix. We have to normalize u1 and u2 : set v2 = u2
u2 1 2
= √
5 1 i.e., λ1 = 2 and λ2 = −18. Associated eigenvectors are , . 1 2
v1 = √
,
5 1 1
1
0
, P =√
−3
5 −2 2
1 . xT A x = −2x2 + 16x1 x2 − 14x2 = 5
1
2
to a quadratic equation in y1 , y2 with no crossproduct term given that λ1 ≥ λ2 . 2
0
P −1 = P DP T
0 −18 is an orthogonal diagonalization of A.
Now set
x = x1
,
x2 2
2
2y1 − 18y2 = −5 2. 2
2
2y1 + 18y2 = −5 3. 2
2
2y1 − 18y2 = 5 correct 4. 2
2
2y1 + 18y2 = 5 x = P y. Then = yT 2
0
2
2
y = 2y1 − 18y2 = 5 .
0 −18 Consequently, when x = P y,
2
2
−2x2 + 16x1 x2 − 14x2 = 2y1 − 18y2 = 5 .
1
2 xT Ax = −2x2 + 16x1 x2 − 14x2
1
2
−2
8 8
−14 x1
.
x2 The eigenvalues λ1 , λ2 of A are the solutions
of
det y1
,
y2 = yT (P T P )D(P T P )y = yT Dy Explanation:
In matrix terms, −2 − λ
8 y = xT A x = (P y)T (P DP T )P y 1. = [x1 x2 ] −1
2 is an orthogonal matrix such that 10.0 points Make an orthogonal change of variables
that reduces 1 −1
v2 = √
,
5 2 1 2
P = [v1 v2 ] = √
5 1 A = P
006 −1
,
2 are thus orthonormal, and is an orthogonal diagonalization of A when
2
0 u2 = and these are orthogonal since λ1 = λ2 . The
normalized eigenvectors Then P = [v1 v2 ] is an orthogonal matrix
and
A = P DP −1 D= 2
,
1 u1 = 1
1
= √
−2
5 u1
u1 v1 = 3 8
−14 − λ 007 10.0 points If AT = A and if vectors u and v satisfy
Au = 3u and Av = 4v, then u · v = 0.
True or False? 1. TRUE correct
2 = λ + 16λ − 36 = (λ − 2)(λ + 18) = 0 , 2. FALSE
Explanation: Version 073 – EXAM04 – gilbert – (56780)
The given vectors u and v are eigenvectors
of A corresponding to eigenvalues λ1 = 3 and
λ2 = 4. But when A is symmetric, then any
two eigenvectors from diﬀerent eigenspaces
are orthogonal. Thus u and v must be orthogonal, that is, u · v = 0.
Consequently, the statement is Now (−1, −2), 2. x-intercept = − (1, −3) . 4
5 3. x-intercept = − (0, 1), 14
15 13
15 1
−2 (0, 1), 4
−2 −2
6 ˆ
x = = 4
−2 1 1
0 1 −2 1 −1 1 0 1 1 1
1 −2
−1 ,
0 1 −2
6 b
m = −1
.
−7 3 −2 .
b = 1 ˆ
x = −1 −1 . −1
−7 −1
−7 x-intercept = − = −3
2 2
.
3 10.0 points If A is symmetric, then the change of variable x = P y transforms
Q(x) = xT A x −3 b
m −2
6 Consequently, the Least Squares Regression line is
3
y = − x−1
2
and so its (1, −3) . The least squares regression line for this data
ˆ
is y = mx + b where x is the solution of the
normal equation
ˆ
AT A x = AT b , −2
.
6 4
−2 1 6 2
20 2 4 009 . , Thus the normal equation is = Explanation:
The design matrix and list of observed values for the data 1
1
A = 1 1
−1
4
−2 = b
m 11
5. x-intercept = −
15 (−1, −2), −1
−7 So 2
4. x-intercept = − correct
3 are given by 1
1 10.0 points 1. x-intercept = − (−2, 3), 1
0 3 −2 1 =
−3 while
AT A = Find the x-intercept of the Least Squares
Regression line y = mx + b that best ﬁts the
data points
(−2, 3), 1
1
−2 −1 AT b = TRUE .
008 4 into a quadratic form with no cross-product
term for any orthogonal matrix P .
True or False?
1. TRUE Version 073 – EXAM04 – gilbert – (56780) i.e., λ1 = 16, λ2 = 1. Eigenvectors x1 and x2
associated with λ1 and λ2 are 2. FALSE correct
Explanation:
When P is orthogonal and x = P y, then
Q(x) = xT A x = (P y)T A (P y)
= yT P T A P y = yT (P T AP )y .
But this will contain cross-product terms unless P T AP is a diagonal matrix, i.e., unless A
is orthogonally diagonalized by P .
Consequently, the statement is
FALSE .
010 2 0
3 2 , A = U ΣV T
of A?
1 −2
2 1
−1
correct
2 1 1
3. V = √
5 2 −2
1 2 −1
1 2 Explanation:
Since
AT A = 2 3
0 2 x2 = −1
;
2 these are orthogonal. Associated orthonormal
eigenvectors are thus
1 2
v1 = √
,
5 1 1 −1
v2 = √
.
5 2 Consequently, one choice for V is
−1
2 . Since multiples of x1 and x2 are again eigenvectors of AT A corresponding respectively to
λ1 = 16 and λ2 = 1, there are other choices
for V . But these are not among the choices
listed.
011 10.0 points If A is an m × n matrix and b is in Rm , the
general least-squares problem is to ﬁnd an x
that makes Ax as close as possible to b.
True or False? 1 2
2. V = √
5 1 4. V = 2
,
1 10.0 points which of the following could be a choice for the
matrix V in a Singular Value Decomposition 1. V = x1 = 1 2
V = [v1 v2 ] = √
5 1 When
A = 5 13 6
2 0
,
=
6 4
3 2 1. TRUE correct
2. FALSE
Explanation:
ˆ
A least squares solution of Ax = b is an x
in Rn such that b − Aˆ ≤ b − Ax for all
x
x in Rn . Note that b − Ax is the distance
from Ax to b, thus the goal is to minimize
that distance.
Consequently, the statement is
TRUE . the eigenvalues of AT A are the solutions of
(13 − λ)(4 − λ) − 36 = λ2 − 17λ + 16
= (λ − 16)(λ − 1) = 0 ,
012 10.0 points Version 073 – EXAM04 – gilbert – (56780) 6
x2 Which one of the following is the graph of
4.
5x2
1 − 4x1 x2 + 2x2
2 = 16 ?
x1 (All axes drawn to the same scale.)
x2
1.
x2 x1
5. correct x1 x2
2.
x2
x1 6. x1 x2
3. Explanation:
In matrix terms,
5x2 − 4x1 x2 + 2x2 = xT Ax
1
2 x1
with x = x1
,
x2 A = 5
−2 −2
2 . But if λ1 , λ2 are the eigenvalues of A and
v1 , v2 are respective corresponding normalized eigenvectors, then by the Principal Axes Version 073 – EXAM04 – gilbert – (56780) 7 theorem,
013
T x Ax = y λ1
0 T 0
2
2
y = λ1 y1 + λ2 y2
λ2 Use an orthogonal matrix P to identify where
P = [v1 v2 ], y1
,
y2 y = x2 + 4xy − 2y 2 = 0
x = Py . Now
det[A − λI] = 5−λ
−2 −2
2−λ as a conic section in x1 , y1 without cross-term
when
x1
x
.
= P
y1
y
2
1. ellipse 2x2 − 3y1 = −1
1 = λ2 − 7λ + 6 = (λ − 6)(λ − 1) ,
so λ1 = 6, λ2 = 1 and
1
2
v1 = √
,
5 −1 10.0 points 2
2. hyperbola 2x2 − 3y1 = −1
1
2
3. ellipse 2x2 + 3y1 = 1
1 1 1
v2 = √
.
5 2 2
4. straight lines 2x2 − 3y1 = 0 correct
1 5. point (0, 0) Thus the graph of 2
6. hyperbola 2x2 + 3y1 = 1
1 5x2 − 4x1 x2 + 2x2 = 16
1
2 Explanation:
The quadratic relation is that of the ellipse
2
2
6y1 + y2 = 16 x2 + 4xy − 2y 2 = 0 with respect to the y1 y2 -axes. Since
x = P y = [y1 y2 ] y1
y2 xT Ax = xT = y1 v1 + y2 v2 , the y1 -axis is the line tv1 , while the y2 -axis is
the line tv2 . Consequently, the graph of the
ellipse is given by
x2 can be written in matrix terms as y2 2
x = 0
−2 x
.
y
To eliminate the cross-term we orthogonally diagonalize A by ﬁnding the eigenvalues
λ1 , λ2 and corresponding normalized eigenvectors v1 , v2 of A. For then A = P DP T
with
where x = P = [v1 v2 ],
x1 1
2 D = λ1
0 0
,
λ2 and
2
xT Ax = yT Dy = λ1 x2 + λ2 y1 ,
1 y1 setting
P y = x, y = x1
,
y1 x = x
.
y Version 073 – EXAM04 – gilbert – (56780)
Since A is symmetric, P will be orthogonal if
λ1 = λ2 . But λ1 and λ2 , then the spectral decomposition of
A is given by
T
A = λ1 v1 v1 + λ2 v2 vT , det[A − λI] = (1 − λ)(−2 − λ) − 4
= λ2 + λ − 6 = (λ − 2)(λ + 3) = 0. and the component determined by λ1 is Thus λ1 = 2 and λ2 = −3, so that in x1 , y1
coordinates the quadratic relation becomes which as a degenerate conic section is the
familiar form of a pair of straight lines. Compute the component of the spectral decomposition determined by λ1 when A is the
symmetric 2 × 2 matrix with eigenvalues
λ2 = −9 and corresponding eigenvectors
2
,
−4 x1 = 1. 9 4
5 −2 1 1
5 −2 4. −
5. − 9 4
5 2 6. 2
.
1 = v2 = 1
√ 2
2 5 −4
x2
x2 1
1
= √
5 −2 1 2
= √
.
5 1 Then
T
v1 v1 = 1 1
1 1 −2
[ 1 −2 ] =
,
5 −2
5 −2 4 T
v2 v2 = 1 2
1 4 2
[2 1] =
.
1
5
5 2 1 Thus
A =
2
4 −2
correct
4 9 1
5 −2 x1
x1 while 1 1 −2
5 2 1 2. −
3. x2 = v1 =
and 10.0 points λ1 = 1, T
λ1 v1 v1 . Now the given eigenvectors x1 , x2 are orthogonal, but not orthonormal, so set 2
2x2 − 3y1 = 0,
1 014 8 −2
4
2
1 1 4 2
5 2 1 Explanation:
When v1 , v2 are orthonormal eigenvectors
of A corresponding to respective eigenvalues λ1
5 1
−2 λ2
−2
+
4
5 4
2 2
.
1 Consequently, the component of the spectral decomposition of A determined by λ1 is
1 1 −2
5 −2 4 . ...

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