MAT 1341A, Fall 2018 The Final Exam
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
MAT 1341A – The Final Exam
Fall 2018
1. F13fs, Q2: Let 于 = {f | f : R → R} denote the vector space of all real-valued functions. Which two of the following are subspaces of 于? (1)
S = {f ∈ 于 | f (一2)f (2) = 0}
T = {f ∈ 于 | f (0) = 0}
U = {f ∈ 于 | f (1) = f (0)}
V = {f ∈ 于 | f (1) = 一1}
mark (X) the correct answer:
S and T
S and U
S and V
T and U
T and V
U and V
: S is not a subspace: Take f(x) = x + 2 and g(x) = x _ 2. Then both f, g e S . We have (f + g)(x) = 2x. So that f + g S .
V is not a subspace: Take f(x) = _x and g(x) = _1. Then both f, g e V but f + g V .
2. F13fs, Q3: Suppose {u, v, w} is a set of vectors in a vector space V .
Which of the following statements is equivalent to (1)
”{u, v, w} is linearly independent”?
I. None of the vectors u, v or w is a linear combination of the other vectors in {u, v, w}.
II. None of the vectors u, v or w is a multiple of any other single vector in {u, v, w}.
III. If a, b, c are scalars then au + bv + cw = 0 implies a = b = c = 0.
IV. If a = b = c = 0, then au + bv + cw = 0.
mark (X) the correct answer:
A I. and II.
I. and III.
I. and IV.
D II. and III.
II. and IV.
F III. and IV.
B : The definition of linearly independent set is equivalent to I. or to III.
3. F17fs, Q3: Let A be a square n × n matrix with n ≥ 2.
Which of the following statements are true? (1)
I. If rank(A) = 1, there is just one parameter in the general solution of the system Ax = 0.
II. If rank(A) = 1, there are n 一 1 parameters in the general solution of the system Ax = 0.
III. If A is invertible, the homogeneous system Ax = 0 has a unique solution.
IV. If the system Ax = 0 has infinitely many solutions, then rank(A) = n.
mark (X) the correct answer:
I. only
II. only
I. and III.
II. and III.
I. and IV.
II. and IV.
: Using the formula rank(A) + #parameters = #columns, we obtain that if rank(A) = 1, then #parameters = n _ 1, so II. is correct. Also IV. is incorrect by the same formula.
(see the last page for the table of trigonometric functions)
mark (X) the correct answer:
^2(cos(一7π/12) + i sin(一7π/12))
^2(cos(11π/12) + i sin(11π/12))
^2(cos(5π/12) + i sin(5π/12))
^2(cos(π/12) + i sin(π/12))
^2(cos(一π/12) + i sin(一π/12))
^2(cos(一5π/12) + i sin(一5π/12))
: We have
= = = ^2ei( 一亓/3 一3亓/4) = ^2ei( 一 13亓/12) = ^2ei(11亓/12) .
5. F11fs, Q7: The dimension of the subspace U = {A ∈ M3×3 (R) | AT = 一A} is (1)
mark (X) the correct answer:
A 0
B 2
C 3
D 4
E 6
F 9
C |
: We need 3 parameters to describe this subspace as
U = { ╱ 0_a · _b |
a 0 _c |
·(、) | a, b, c e R}. |
mark (X) the correct answer:
A is not invertible.
The third row of A_1 is (一1, 一1, 1).
The second row of A_1 is (1, 2, 一1).
The first row of A_1 is (2, 0, 一1).
The second column of A_1 is (0, 2, 一1)T .
All of B, C, D, E are true.
: The third row of A is a sum of the first two. So the rows are linearly dependent, hence, A is not invertible
7. F16fs, Q7: The set {u, v, w} is an orthogonal set of vectors, where
u = (0, 3, 4), v = (1, 0, 0) and w = (0, 4, 一3).
If (0, 一1, 一1) = au + bv + cw, then (a, b, c) =
mark (X) the correct answer:
(一 , 0, 一 )
(一 , 0, 一 )
( , 0, )
(一 , 0, )
(一7, 0, 一1)
(0, 一1, 一1)
a1 = (0 ;3 ;4.2(42(;一)1 ; 一 1) = _ , a2 = 0, a3 = (0 ;4 ; 一3(3).一21 ; 一 1) = _ .
8. F16fs, Q9: Consider the matrix A = ╱0(3) 3(1)、. Answer the following questions (Yes/No):
(1)
. Is 3 the only eigenvalue of A?
. Is the dimension of the eigenspace corresponding to eigenvalue 3 equal to 1? . Is A diagonalizable?
mark (X) the correct answer:
Yes, Yes, Yes
Yes, Yes, No
No, Yes, Yes
No, Yes, No
No, No, Yes
Yes, No, No
: The eigenvalue is 3 and the dimension is 1.
9. F14fs, Q9: Let A = !(、) for some numbers a, b, c, d, e, f, g, h, i.
If det(A) = 3, find the determinant of the matrix ﹔ 一(一) 2(2)b(a)、
( 4i c f 一 2c!.
mark (X) the correct answer:
6
-6
12
-12
24
-24
: perform the row and column operations to reduce it to A.
10. F14fs, Q10: Let 于 = {f | f : R → R} denote the vector space of real-valued functions. Which of the following are linearly independent subsets in 于? (1)
S = {sin x, cos x}
T = {1, sin x, cos x}
U = {1, sin2 x, cos2 x}
V = {1, 2 sin2 x, 3 cos2 x}
mark (X) the correct answer:
S and T
S and U
S and V
T and U
T and V
U and V
: We have sin2 x + cos2 x = 1.
11. F13fs, Q11: (3 points) Let k ∈ R and consider the linear system in the unknowns x, y, z:
『 kx + 2y + z = 0
│y + z = 0
『(3z = 0
Find all values of k for which this system has
(a) a unique solution,
(b) infinitely many solutions,
(c) no solutions.
Solution:
As this is a homogeneous system it always has a solution.
There are no such k’s |
We compute
det A = det !(、) = 3k .
The the system has a unique solution if and only if A is invertible if and only if det A 0. So
For (b) k = 0 |
12. F14fs, Q12: (3 points) Let v1 = (1, 0, 0, 一1), v2 = (1, 一1, 0, 0), v3 = (1, 0, 1, 0). Consider the subspace U = Span{v1 , v2 , v3 } ∈ R4 . (a) Explain (you may refer to results learned in class) why {v1 , v2 , v3 } is a basis of U . (1) Solution: It is enough to show that {v1 , v2 , v3 } are linearly independent. We have rank 0一01 一001!(、) = rank 0一01 一111!(、) = 3 which is the same as the number of vectors. (b) Use the Gram-Schmidt algorithm to find an orthogonal basis for U . (1) |
The orthogonal basis is:
write it as row vectors |
Solution:
Set u1 = v1 ,
u2 = v2 一 u1 = (1, 一1, 0, 0) 一 (1, 0, 0, 一1) = ( , 一1, 0, ).
u3 = v3 一 u1 一 u2 =
(1, 0, 1, 0) 一 (1, 0, 0, 一1) 一 ( , 一1, 0, ) =
(1, 0, 1, 0) 一 ( , 0, 0, 一 ) 一 ( , 一 , 0, ) = ( , , 1, )
Hence, {(1, 0, 0, 一1), ( , 一1, 0, ), ( , , 1, )} is an orthogonal basis of U .
(c) Find the best approximation by a vector in U to the vector (0, 1, 1, 1). (1)
|
The best approximation is: write it as a row vector |
Solution:
pTojU (0, 1, 1, 1) =
(1, 0, 0, 一1) + ( , 一1, 0, ) + ( , , 1, ) =
(1, 0, 0, 一1) + ( , 一1, 0, ) + ( , , 1, ) =
一 (1, 0, 0, 一1) 一 ( , 一1, 0, ) + ( , , 1, ) = (一 , , , ) = (一1, 3, 5, 3).
13. F15fs, Q13: (4 points) Let A = ﹔ 1(1) 3(0) 一12、
(1 0 4 !.
(a) Compute and factorize the characteristic polynomial of A to show that the only eigenvalues of A are 2 and 3.
The polynomial is:
det(A 一 λI) = det 1 一11 λ 3 0一0 λ 4一1一2 (3 一 λ) det ╱ 11(一) λ 4一(一)2λ、=
(3 一 λ) │(1 一 λ)(4 一 λ) + 2/ = (3 一 λ)(4 一 5λ + λ2 + 2) =
(3 一 λ)(λ 一 3)(λ 一 2) = 一(λ 一 3)2 (λ 一 2).
(b) Find a basis of the eigenspace corresponding to the eigenvalue 2. (1)
E2 = Null(A 一 2I) = Null ﹔ 一11 1(0)
( 1 0
So that {(一2, 1, 1)} is a basis for E2 .
2 ! = Null (0 0 0 ! = {(一2s, s, s) | s ∈ R}
(c) Find a basis of the eigenspace corresponding to the eigenvalue 3. (1)
E3 = Null(A 一 3I) = Null 一112 一112!(、) = Null !(、) = {(一t, s, t) | s, t ∈ R}
So that {(一1, 0, 1), (0, 1, 0)} is a basis for E3 .
(d) Find an invertible matrix P and a diagonal matrix D such that P_1 AP = D . (1)
P =
D =
Solution:
Columns of the matrix P consist of bases of the respective eigenspaces P = ﹔ 一12 一01
2023-07-15