18.06
FINAL EXAM
May 24, 2007
Your PRINTED name is: SOLUTIONS
Please circle your recitation:
(1) (2) (3) M2 M3 M3 2-131 A. Osorno 2-131 A. Osorno 2-132 A. Pissarra Pires
Grading 1 2 3 4 5 6 7 8 Total:
(4) T 11 2-132 K. Meszaros (5) T 12 2-132 K. Meszaros
18.06 Problem Set 5 - Solutions
Due Wednesday, March 21, 2007 at 4:00 p.m. in 2-106
Problem 1 Wednesday 3/14
Do problem 10 of section 4.3 in your book.
Solution 1
1
1
1
1
1
0
1
1
0
1
3
4
0
1
3
4
0
1
9
16
0
1
9
16
C
0
0
1 D 8
=
27 E 8
20
F
64
0
0
1
1
8
18.06 Problem Set 6 - Solutions
Due Wednesday, April 11, 2007 at 4:00 p.m. in 2-106
Problem 1 Wednesday 4/4
Do problem 9 of section 6.1 in your book.
Solution 1
( a) Multiply A on the left to both sides of the equation Ax = x to get AAx = Ax. But
AAx = A2
18.06 Problem Set 10 - Solutions
Due Friday, May 11, 2007 at 1:00 p.m. in 2-106
Problem 1 Wednesday 5/2
Do problem 7 of section 7.1 in your book.
Solution 1
(a) T (T (v) = v, transformation T 2 is linear.
(b) T (T (v) = T (v + (1, 1) = v + (2, 2), transfo
SOLUTIONS TO THE PRACTICE SET
Problem 1. We know (section 6.4) that is a real diagonal matrix. Thus H = ,
and AH = (U H U)H = U H H (U H )H = A. So A is hermetian and hence has real eigenvalues (section 10.2). Further, since U H = U 1 , A must have the sa
18.06 FINAL SOLUTIONS
a b a+b
Problem 1. (10 points) B = b c b + c . We know that symmetric matrices have
x y
z
real eigenvalues and orthogonal eigenvectors. So we set x = a + b and y = b + c. This
leaves only the singularity of B. For this, we note that
SOLUTIONS TO PSET 3
Problem 1. a) (5 points) Choose the lattice consisting of points (x, y) such that x and y
are integers.
b) (5 points) Well, the fact that cx stays in the set says that the set is a union of lines. So
take the x-axis union the y-axis.
P
SOLUTIONS TO PSET 5
1. Well, AA1 = I means: if A1 , ., An denote the rows of A, and B1 , ., Bn denote the
columns of A1 , then Ai B j = i j where the symbol i j is equal to 1 if i = j and 0 if i = j.
But this says that B1 is orthogonal to the space spanne
SOLUTIONS TO PROBLEM SET 2
Problem 1. 1) (2.5 points) AB = A B1 B2 B3 B4 = AB1 AB2 AB3 AB4 .
A1
A1 B
2) (2.5 points) AB =
2 B = A2 B .
A
A1
A1 B1 A1 B2 A1 B3 A1 B4
B1 B2 B3 B4 =
.
3) (2.5 points) AB =
2
A
A2 B1 A2 B2 A2 B3 A2 B4
1
B
4) (2.5 points) AB =
SOLUTIONS TO PSET 4
1 2 1
1 2 1
1 2 1
1. a) (5 points) 2 6 3 0 2 1 0 2 1 so this matrix is invert0 2 5
0 2 5
0 0 4
ible. Therefore, the column space consists of all vectors in R3 , and no nontrivial linear
combo. of the rows can be zero.
1 1 1 b1
1
SOLUTIONS TO PSET 6
Problem 1. (5 points each) a) 1 = det(I) = det(Qt Q) = det(Qt )det(Q) = (det(Q)2 ,
so det(Q) = 1.
b) Suppose that |det(Q)| > 1. Then we have that, for each m, |det(Qm )| = |det(Q)|m ,
which goes to as m . But Qm is an orthogonal matrix
SOLUTIONS TO PSET 7
1. (5 points each)
2
= ( )(3 ) 4 = 2 3 4 = ( 4)( +
2 3
1); so the eigenvalues are 4 and 1. So, to get the eigenvector corresponding to 4,
4 2
x1
1
we solve
= 0 and arrive at
. For the 1 eigenvector, we solve
2 1
x2
2
1 2
x1
2
= 0 and g
SOLUTIONS TO PSET 9
Problem 1. (5 points each)
1. B = M 1 AM and C = N 1 BN imply C = N 1 M 1 AMN = (MN)1 A(MN). If B is
similar to A and C is similar to B, then A is similar to C.
2. F 1 AF = C = G1 BG, so B = GF 1 AFG1 = (FG1 )1 A(FG1 ). If C is similar
SOLUTIONS TO PSET 8
Problem 1. (5 points each)
1. We carry out the three steps on page 306. Firstly the eigenvalues of
clearly 4 and 1. Thus, the eigenvectors can be found by solving
3
0
3
0
1
0
0 3
0 3
4
0
x1
x2
3
1
are
= 0 and
1
5
1
1
. Now, u(0) =
=3
+
18.06 Problem Set 7 - Solutions
Due Wednesday, April 18, 2007 at 4:00 p.m. in 2-106
Problem 1 Wednesday 4/11
Do problem 9 of section 8.3 in your book.
Solution 1
We get that:
u1 = P u0 = (0, 0, 1, 0),
u2 = P u1 = (0, 1, 0, 0),
u3 = P u2 = (1, 0, 0, 0),
u4
18.06 Problem Set 8 - Solutions
Due Wednesday, April 25, 2007 at 4:00 p.m. in 2-106
Problem 1 Wednesday 4/18
Do problem 5 of section 6.3 in your book.
Solution 1
To show v + w is constant we need to show
d
dt (v
+ w) = 0. We have
d
dv dw
(v + w) =
+
=wv+v
18.06 Problem Set 4 - Solutions
Due Wednesday, Mar. 14, 2007 at 4:00 p.m. in 2-106
Problem 1 Wednesday 2/28
Do problem 37 of section 3.5 in your book.
Solution 1
We want to nd a basis for the space of polynomials of degree 3, i.e., the space that contains
18.06 Problem Set 1 - Solutions
Due Wednesday, 12 September 2007 at 4 pm in 2-106.
Problem 0: from the book.(50=10+10+10+10+10)
(a) problem set 1.2, problem 8.
(a) F.
A Counterexample: u = (1, 0, 0), v = (0, 1, 0) and w = (0, 1, 1).
(b) T.
Proof: u (v + 2
18.06 Problem Set 2 - Solutions
Due Wednesday, 19 September 2007 at 4 pm in 2-106.
Problem 1: (10=5+5) (a) Do problem 12 from section 2.4 (P67) in your book.
First AB =
a 0
, BA =
c 0
0 a
=
0 c
b = c = 0 implies a = d.
a 0
Thus A =
=a
0 a
(b) Do problem 3
18.06 Problem Set 3 - Solutions
Due Wednesday, 26 September 2007 at 4 pm in 2-106.
Problem 1: (10=2+2+2+2+2) A vector space is by denition a nonempty set V
(whose elements are called vectors) together with rules of addition (u, v V
u + v V ) and scalar m
18.06 Problem Set 4 - Solutions
Due Wednesday, 10 October 2007 at 4 pm in 2-106.
Problem 1: (10=2+2+2+2+2) Decide whether the following set of vectors are
linearly dependent or independent. (Give reasons)
(a) (1, 2, 3), (2, 3, 1), (3, 1, 2).
Solution Line
18.06 Problem Set 6 - Solutions
Due Wednesday, 24 October 2007 at 4 pm in 2-106.
Problem 1: (10=3+3+4) Do problem 4 from section 4.4 (P 228) in your book.
1 0
Solution (a) Example: Q = 0 1, then it has orthonormal columns but
0 0
1 0 0
QQT = 0 1 0 = I.
0
18.06 Problem Set 7 - Solutions
Due Wednesday, 07 November 2007 at 4 pm in 2-106.
1 3
1 1
3 5
1 1
Problem 1: (12=3+3+3+3) Consider the matrix A =
10 10 10 14 .
4 4 4 8
(a) If one eigenvector is v1 = 1 1 0 0
Solution Av1 = 2 2 0 0
T
T
, nd its eigenvalue
18.06 Problem Set 8 - Solutions
Due Wednesday, 14 November 2007 at 4 pm in 2-106.
0.8 0.3
.
0.2 0.7
(a) Check that A is a positive Markov matrix, and nd its steady state.
Problem 1: (20=5+5+5+5) Consider the matrix A =
Solution A is obviously positive Mar
18.06 Problem Set 5 - Solutions
Due Wednesday, 17 October 2007 at 4 pm in 2-106.
Problem 1: (10) Do problem 22 from section 4.1 (P 193) in your book.
Solution The equation x1 + x2 + x3 + x4 = 0 can be rewritten in the matrix form
x1
x2
1 1 1 1 = 0.
x3
18.06 Problem Set 9 - Solutions
Due Wednesday, 21 November 2007 at 4 pm in 2-106.
Problem 1: (15) When A = SS 1 is a real-symmetric (or Hermitian) matrix, its
eigenvectors can be chosen orthonormal and hence S = Q is orthogonal (or unitary).
Thus, A = QQT
18.06 Problem Set 10 - Solutions
Due Thursday, 29 November 2007 at 4 pm in 2-106.
Problem 1: (15=5+5+5) Take any matrix A of the form A = B H CB, where B has full
column rank and C is Hermitian and positive-denite.
(a) Show that A is Hermitian.
Solution