Math 2310 HW7 due in class March 19, 2014
1. Problems 3, 6, 18, 24 in section 3.6 of the text.
2. Let v be a nonzero vector in Rn and A an n n matrix.
(a) True or False: It is possible for v to be in the row space and null space of A. If true, give an
exa
Math 2310 Homework 5
Due: October 11 (beginning of class)
Read:
Read 3.2, 3.3, and 3.4 from the textbook. I recommend that you attempt
to solve the Worked Examples before looking at the provided solutions.
Turn in:
3.2: 1, 2, 3, 4, 5, 6, 7, 8, 30, 37.
3.3
Math 2310 Homework 6
Due: October 25 (beginning of class)
Read:
Read 3.5, 3.6, and 4.1 from the textbook. I recommend that you attempt
to solve the Worked Examples before looking at the provided solutions.
Turn in:
3.5: 2, 13, 15, 22, 26, 43, 45.
3.6: 2,
Math 2310 Homework 7
Due: November 1 (beginning of class)
Read:
Read 4.2, 4.3, and 4.4 from the textbook. I recommend that you attempt
to solve the Worked Examples before looking at the provided solutions.
Turn in:
4.2: 1, 11, 15, 17, 21, 22.
4.3: 12, 17,
7.1.23) If not, then A must have some other eigenvalue = 0 with corresponding eigenvector v. Then we can compute
Ak v = (Ak1 )Av = (Ak1 )v = Ak1 v = = k v = 0.
But if Ak = O, then Ak v = Ov = 0, a contradiction. Hence, cannot be an
eigenvalue of A.
7.2.23
6.1.1) a) In order to be a linear transformation, we should have L(0) =
L(00) = 0L(0) = 0. However, this gives us L([0 0]) = [1 0 0] = 0, so it is not a
linear transformation.
b) Let [u1 u2 ] and [v1 v2 ] be arbitrary vectors in R2 , and let c be an arbit
5.3.21) We can compute
=
=
=
=
=
=
=
=
=
=
1
1
|u + v|2 |u v|2
4
4
1
(u + v, u + v) (u v, u v)
4
1
(u, u + v) + (v, u + v) (u, u v) (v, u v)
4
1
(u, u + v) + (v, u + v) (u, u v) + (v, u v)
4
1
(u + v, u) + (u + v, v) (u v, u) + (u v, v)
4
1
(u + v (u v),
4.9.34) a) The rank is the dimension of the row space, which is spanned by
three vectors, so it is at most 3. One can easily have a matrix A = cfw_aij of rank
3 by setting aij = 1 if i = j and aij = 0 otherwise.
b) The row space is the span of four vecto
4.6.16)
10
0 0
00
One possibility is
0
000
0
0 , 0 1 0 , 0
0
000
0
0
0
0 , 1
1
0
0
0
0
0
1
1
0
4.6.33) The span of
0 and 0
0
0
1
0
0
0
0
0 , 0
0
1
0
0
0
1
0
0 , 0
0
0
is one possibility.
4.6.42) Since dim W = dim V , W is also nite dimensional. Let dim
4.3.1) Yes, it is a subspace. It is clearly a subset of R2 . To show that it is
a subspace, Theorem 4.3 says that we need only check that it is closed under
addition and scalar multiplication. For the former if (x, x) and (y, y ) are two
arbitrary points
3.4.14) By Theorem 3.12, A(adj(A) = det(A)In . Both sides are n n
matrices, so we can take the determinant of both to get det(A(adj(A) =
det(det(A)In ). The right hand side of this is a scalar matrix with det(A) in each
of the n entries on the diagonal. T
3.2.4) By Theorem 3.6, adding a constant multiple of one column to another
doesnt change the determinant, so the determinant of the new matrix is still
-2.
3.2.8) Yes, det(AB ) = det(A) det(B ) = det(B ) det(A) = det(BA). The
equalities on the end follow
2.3.25) Suppose that AB is nonsingular. By Theorem 2.9, the only solution
to AB x = 0 is x = 0. If B were singular, then there would be some y = 0 such
that B y = 0, in which case, we would have AB y = A(B y) = A0 = 0, which
gives us a nontrivial solution
Math 2310 Homework 4
Due: October 4 (beginning of class)
Read:
Read 2.6, 2.7, and 3.1 from the textbook. I recommend that you attempt
to solve the Worked Examples before looking at the provided solutions.
Turn in:
2.6: 5,9,12,15,18.
2.7: 2,9,12,16,17,22,3
Math 2310 Homework 3
Due: September 27 (beginning of class)
Read:
Read 2.2, 2.3, and 2.5 from the textbook. I recommend that you attempt
to solve the Worked Examples before looking at the provided solutions.
Turn in:
2.2: 8,11,12.
2.3: 3,8,19,27,31.
2.5:
Math 2310 Prelim 1, Feb. 26, 2014
Name:
Work + Communication = Credit
Problems 1,3,4, were worth 20 points. Problem 2 was worth 25 points and problem 5 was worth 15 points.
1. Here are three linear equations in three variables:
x + 2y z
y + 3z
2x + 5y + z
Math 2310 Prelim 1, Feb. 26, 2014
Name:
Work + Communication = Credit
Problems 1,3,4, were worth 20 points. Problem 2 was worth 25 points and problem 5 was worth 15 points.
1. Here are three linear equations in three variables:
2x + 5y + z
y + 3z
x + 2y z
Math 2310
Cornell University. Ithaca, NY.
Solutions to Problem Set #2.
You can nd here the solutions to the problems whose solution is not on the back of the
book.
Problem 1.2.26
Solution:
The vectors w = (x, y ) with (1, 2) w = x + 2y = 5 lie on a line i
Math 2310
Cornell University. Ithaca, NY.
Solutions to Problem Set #1.
You can nd here the solutions to the problems whose solution is not on the back of the
book.
Problem 1.1.15
Solution:
3
The point 4 v + 1 w is three-fourths of the way to v starting fr
Math 2310
Cornell University. Ithaca, NY.
Solutions to Problem Set #5.
Problem 3.2.9 A lot of people had trouble with the last question, so please read the solution.
1. False (Any singular square matrix has free variables).
2. True (An invertible square m
Math 2310
Cornell University. Ithaca, NY.
Solutions to Problem Set #3.
You can nd here the solutions to the problems whose solution is not on the back of the
book.
Problem 2.2.5
Solution:
6x + 4y is 2 times 3x + 2y . There is no solution unless the right
Math 2310
Cornell University. Ithaca, NY.
Solutions to Problem Set #4.
You can nd here the solutions to the problems whose solution is not on the back of the
book.
Problem 2.5.2
Solution:
A row exchange gives P 2 = I , and hence P 1 = P . Hence P 1 =
001
Math 2310
Cornell University. Ithaca, NY.
Solutions to Problem Set #6.
Problem 3.4.1 A lot of people had trouble with the last question, so please read the solution.
Solution:
2 4 6 4 b1
24
6
4
b1
2464
b1
2 5 7 6 b2 0 1
1
2 b2 b1 0 1 1 2
b 2 b1
2 3 5 2 b
Very short solutions to extra problems in Sections 3.1 and Section 3.2 which are not in the
text
Section 3.1
1
2
3
12 - 1 , 2 are in P , but their sum, 3 is not.
2
0
2
17 - (a) I and I are invertible. However, their sum is the zero matrix which is not i
Math 2310 Homework 1
Due: September 11 (beginning of class)
Read:
Read 1.1 and 1.2 from the textbook. I recommend that you attempt to
solve the Worked Examples before looking at the provided solutions.
Turn in:
1.1: 1,3,5,13,19,26,28,31.
1.2: 4,6,8,12,16,
Math 2310 Homework 2
Due: September 20 (beginning of class)
Read:
Read 1.3, 2.1, and 2.4 from the textbook. I recommend that you attempt
to solve the Worked Examples before looking at the provided solutions.
Turn in:
1.3: 1,8,14.
2.1: 5,8,12,31,35.
2.4: 1
cos(k) sin(k)
. This holds because if A rotates every
sin(k) cos(k)
vector counterclockwise by an angle of , then applying A k times means that
in total, you rotate it by an angle of k. (This isnt really a solution to the
problem as stated, but it does e
1.1.18) A homogeneous linear system means that the constants on the right
side are all zero. If all of the variables are zero, then the left side of each
equation is also zero, so this is a solution. Because there is a solution, the
system is consistent.
1.
Let A =
2
2
2
1
.
Compute the eigenvalues and eigenspaces of A.
We start by computing the characteristic polynomial of A as
det (A I )
=
det
2
2
2
1
= (2 )(1 ) (2)(2)
= 2 + 2 4
= 2 + 6
=
( 2)( + 3)
The eigenvalues are the roots of the characteristic p