**Unformatted text preview: **Version 055 – EXAM04 – gilbert – (56780)
This print-out should have 14 questions.
Multiple-choice questions may continue on
the next column or page – ﬁnd all choices
before answering.
001 10.0 points Compute the component of the spectral decomposition determined by λ1 when A is the
symmetric 2 × 2 matrix with eigenvalues
λ1 = −4, λ2 = 11 and corresponding eigenvectors
2
,
−4 x1 =
1. 11 1
5 −2 x2 = 2
.
1 4 1
2. −
5 −2 Then
1 1
1 1 −2
[ 1 −2 ] =
,
5 −2
5 −2 4 T
v1 v1 = while Thus
λ1
5 002 5. − 4 1
5 2 6. − 4 4
5 2 λ2
−2
+
4
5 4
2 2
.
1 4 1
5 −2 −2
4 . 10.0 points If A is symmetric, then the change of variable x = P y transforms −2
1
2
1 Q(x) = xT A x
into a quadratic form with no cross-product
term for any orthogonal matrix P . Explanation:
When v1 , v2 are orthonormal eigenvectors
of A corresponding to respective eigenvalues
λ1 and λ2 , then the spectral decomposition of
A is given by
T
A = λ1 v1 v1 + λ2 v2 vT , and the component determined by λ1 is Now the given eigenvectors x1 , x2 are orthogonal, but not orthonormal, so set
1
√ 2
=
2 5 −4 True or False?
1. TRUE
2. FALSE correct
Explanation:
When P is orthogonal and x = P y, then
Q(x) = xT A x = (P y)T A (P y) T
λ1 v1 v1 . v1 = 1
−2 − 11 4 2
5 2 1 x1
x1 1 2
1 4 2
[2 1] =
.
5 1
5 2 1 T
v2 v2 = 2
4 4. 1 2
= √
.
5 1 x2
x2 v2 = Consequently, the component of the spectral decomposition of A determined by λ1 is −2
correct
4 11 4
3.
5 −2 and A = −2
4 1 1
1
= √
5 −2 = 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 . Version 055 – EXAM04 – gilbert – (56780)
is an orthogonal matrix such that Consequently, the statement is
FALSE .
003 2 1
0
P −1 = P DP T
0 −19 A = P 10.0 points is an orthogonal diagonalization of A. Make an orthogonal change of variables
that reduces Now set xT A x = −3x2 + 16x1 x2 − 15x2 = −5
1
2 x = to a quadratic equation in y1 , y2 with no crossproduct term given that λ1 ≥ λ2 .
1. 2
2
y1 + 19y2 = 5 2. 2
2
y1 − 19y2 = 5 3. 2
2
y1 − 19y2 = −5 correct 4. 2
2
y1 + 19y2 = −5 x Ax = −3x2
1 = [x1 x2 ] −3 − λ
8 Then = yT 1
0
2
2
y = y1 − 19y2 = −5 .
0 −19 + 16x1 x2 −
−3
8 15x2
2 8
−15 2
2
−3x2 + 16x1 x2 − 15x2 = y1 − 19y2 = −5 .
1
2 x1
.
x2 = λ2 + 18λ − 19 004 10.0 points Find the x-intercept of the Least Squares
Regression line y = mx + b that best ﬁts the
data points
(−2, −2), (−1, 1), = (λ − 1)(λ + 19) = 0 ,
i.e., λ1 = 1 and λ2 = −19. Associated eigenvectors are
u1 = u2 = −1
,
2 and these are orthogonal since λ1 = λ2 . The
normalized eigenvectors
1 2
,
v1 = √
5 1 x = P y. Consequently, when x = P y, 8
−15 − λ 2
,
1 y1
,
y2 = yT (P T P )D(P T P )y = yT Dy The eigenvalues λ1 , λ2 of A are the solutions
of
det y = xT A x = (P y)T (P DP T )P y Explanation:
In matrix terms,
T x1
,
x2 1 −1
v2 = √
,
5 2 are thus orthonormal, and
1 2 −1
P = [v1 v2 ] = √
5 1 2 1. x-intercept = − 14
5 2. x-intercept = − (0, 2), 12
5 3. x-intercept = −2 correct
4. x-intercept = − 11
5 5. x-intercept = − 13
5 Explanation: (1, −4) . Version 055 – EXAM04 – gilbert – (56780)
The design matrix and list of observed values for the data
(−2, −2), (−1, 1), are given by (0, 2), and so its
x-intercept = −2 . (1, −4) . 005 1
1
A = 1
1 −2 1 b = .
2
−4 −2
−1 ,
0
1 ˆ
AT A x = AT b , b
m ˆ
x = 10.0 points A vector space V is a subspace of itself.
True or False?
1. FALSE The least squares regression line for this data
ˆ
is y = mx + b where x is the solution of the
normal equation 2. TRUE correct
Explanation:
A set H is a subspace of a vector space V
when . (i) H contains the zero vector 0, Now (ii) the sum u + v of any u, v in H is in H,
1
−2 AT b = 1
−1 1
0 1
1 while −2 1 2 =
−4
−3
−1 , (iii) the scalar multiple cu of any scalar c
and any u in H is in H.
Since V has these three properties, V is a
subspace of itself.
Consequently, the statement is 1
−2 AT A = 1
−1
4
−2 = 1 1
0 1
−2
.
6 1 −2 1 −1 1 0 1 1 TRUE . 006 −2
6 ˆ
x = 4 −2
−2 6 10.0 points When Thus the normal equation is
4
−2 3 u1 =
b
m = −3
.
−1 1
,
−2 u2 = 2
,
1 are eigenvectors of a symmetric 2 × 2 matrix
A corresponding to eigenvalues So
b
m = = 4
−2 1 6
20 2 −2
6 −1 2
4 −3
−1 λ1 = 2 , −3
−1
= −1 −1
2 λ2 = −3 , ﬁnd matrices D and P in an orthogonal diagonalization of A.
. Consequently, the Least Squares Regression line is
1
y = − x−1
2 1. D = 2. D = correct −2
2 0
, P =
1
0 −3
2 0
,
0 −3 1
2 1
1
P = √
−2
5 2
1 Version 055 – EXAM04 – gilbert – (56780) Explanation:
Note that for B to be symmetric, B T = B.
Also note that all diagonal matrices are by
deﬁnition symmetric. Consider B T . 1 2
2 0
, P =
3. D =
−2 1
0 −3
1 2
0
, P =
−2 1
2 4. D = −3
0 5. D = 1 −2
2 0
, P =√
0 −3
5 1 1
2 6. D = −3
0 1
0
1
, P =√
2
5 −2 2
1 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 . Explanation:
When
D= 008
0
, Q = [u1 u2 ],
λ λ1
0 then Q has orthogonal columns and A =
QDQ−1 is a diagonalization of A, but it is
not an orthogonal diagonalization because Q
is not an orthogonal matrix. We have to normalize u1 and u2 : set
1
1
= √
5 −2 u1
u1 v1 = u2
u2 v2 = 1 2
= √
5 1 , is an orthogonal diagonalization of A when
2
0 1
1
0
, P =√
−3
5 −2
007 10.0 points Determine the singular value σ1 for the matrix 3 0
A = 0 −3 .
1 1
√
1. σ1 = 7
√
2. σ1 = 10
√
3. σ1 = 11 correct
4. σ1 = 3 . Then P = [v1 v2 ] is an orthogonal matrix
and
A = P DP −1 D= 4 2
1 . √
5. σ1 = 2 2
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
AT A = 10.0 points If B = P DP
with P an orthogonal matrix and D a diagonal matrix, then B is a
symmetric matrix. 3
0 0
−3 −1 True or False?
1. FALSE
2. TRUE correct = 3
1 0
1
1 10 1
.
1 10 0
−3 1 But then
det(AT A − λI) = 10 − λ
1 1
10 − λ = λ2 − 20λ + 99 = (λ − 11)(λ − 9) . Version 055 – EXAM04 – gilbert – (56780)
Consequently,
σ1 = √ 11 . 5 Explanation:
The normal equations for a least-squares
solution of Ax = b are by deﬁnition
AT Ax = AT b. 009 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. Now,
AT A = True or False?
= 1. TRUE correct
2. FALSE Consequently, the statement is
TRUE . 010 10.0 points Find the least-squares solution of Ax = b
when 0 −1
−4
A = −1 −1 , b = −7 .
1
0
5
1 −21
1.
3 −8
12
2.
11
−4
−7 4. 6
7 5. 0
1 −1
0
1 −1
−1 2 1
1 2 −1
−1 0 and Explanation:
ˆ
A least squares solution of Ax = b is an x
n
in R 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. 3. 0
−1 1 13
correct
3 10 AT b = 0
−1 −1
−1 −4
12
1 .
−7 =
11
0
5 Hence the least squares solution of Ax = b is
the solution x to the equation
12
2 1
.
x=
11
1 2
This can be solved with row reduction or inverse matrices to determine that the solution
is
1 2 −1
3 −1 2
1 13
.
=
3 10 (AT A)−1 (AT b) = 12
11 Consequently, the least squares solution to
Ax = b is
1 13
3 10 011 . 10.0 points Which one of the following is the graph of
5x2 − 4x1 x2 + 2x2 = 16 ?
1
2
(All axes drawn to the same scale.) Version 055 – EXAM04 – gilbert – (56780) 6 x2 x2 1. 5. x1 x1 x2 x2
6. 2. correct x1 x1 x2 Explanation:
In matrix terms, 3. 5x2 − 4x1 x2 + 2x2 = xT Ax
1
2 with
x1 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
theorem,
x2 xT Ax = yT 4. λ1
0 0
2
2
y = λ1 y1 + λ2 y2
λ2 where
x1 P = [v1 v2 ], y = y1
,
y2 x = Py. Now
det[A − λI] = 5−λ
−2
−2
2−λ = λ2 − 7λ + 6 = (λ − 6)(λ − 1) , Version 055 – EXAM04 – gilbert – (56780)
so λ1 = 6, λ2 = 1 and
1
2
v1 = √
,
−1
5 1 1
v2 = √
.
5 2 are orthogonal. Thus u and v must be orthogonal, that is, u · v = 0.
Consequently, the statement is
TRUE . Thus the graph of
5x2 − 4x1 x2 + 2x2 = 16
1
2 013 is that of the ellipse
2
6y1 10.0 points When + 2
y2 A = = 16 with respect to the y1 y2 -axes. Since
x = P y = [y1 y2 ] y1
y2 x2 y2 1 2 −1
1. V = √
correct
5 1 2
2
1 −1
2 1
2 −2
1 1 1 −2
4. V = √
5 2 1
Explanation:
Since
AT A = 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. FALSE
2. TRUE correct
Explanation:
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 , of A? 3. V = y1 3 2 A = U ΣV T 2. V = x1 2 0 which of the following could be a choice for the
matrix V in a Singular Value Decomposition = 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 012 7 2 3
0 2 13 6
2 0
,
=
6 4
3 2 the eigenvalues of AT A are the solutions of
(13 − λ)(4 − λ) − 36 = λ2 − 17λ + 16
= (λ − 16)(λ − 1) = 0 ,
i.e., λ1 = 16, λ2 = 1. Eigenvectors x1 and x2
associated with λ1 and λ2 are
x1 = 2
,
1 x2 = −1
;
2 these are orthogonal. Associated orthonormal
eigenvectors are thus
1 2
v1 = √
,
5 1 1 −1
v2 = √
.
5 2 Version 055 – EXAM04 – gilbert – (56780)
and Consequently, one choice for V is
1 2
V = [v1 v2 ] = √
5 1 −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.
014 10.0 points Use an orthogonal matrix P to identify
3x2 − 4xy + 6y 2 = 6
as a conic section in x1 , y1 without cross-term
when
x1
x
.
= P
y1
y
2
1. ellipse 2x2 + 7y1 = 6 correct
1
2
2. straight lines 2x2 + 7y1 = 0
1
2
3. ellipse 2x2 − 7y1 = 5
1
2
4. hyperbola 2x2 + 7y1 = 6
1 5. point (0, 0)
2
6. hyperbola 2x2 − 7y1 = 5
1 Explanation:
The quadratic relation 3x2 − 4xy + 6y 2 = 6
can be written in matrix terms as
xT Ax = xT 3
−2 −2
x = 6
6 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
λ1 0
,
P = [v1 v2 ], D =
0 λ2 where x = 8 2
xT Ax = yT Dy = λ1 x2 + λ2 y1 ,
1 setting
P y = x, y = x1
,
y1 x = x
.
y Since A is symmetric, P will be orthogonal if
λ1 = λ2 . But
det[A − λI] = (3 − λ)(6 − λ) − 4
= λ2 − 9λ + 14 = (λ − 2)(λ − 7) = 0.
Thus λ1 = 2 and λ2 = 7, so that in x1 , y1
coordinates the quadratic relation becomes
2
2x2 + 7y1 = 6,
1 which as a conic section is the familiar form
of an ellipse. ...

View
Full Document