End of term problems
1. If X and Y are column vectors, then the dot product is given by X Y = X t Y . (The transpose makes
the column matrix X into a row matrix X t , and then multiplying that row matrix by the column matrix
Y gives a 1 1 matrix X t Y , w
Problem Set 5 Part I Solutions
Part I
1. (a) Find all vectors X in R4 that are orthogonal to (1, 1, 1, 1) and (1, 2, 3, 4).
If X = (a, b, c, d), then orthogonality means that its dot product with the two given vectors must
be zero, so
a+b+c+d=0
a + 2b + 3
Problem Set 6 Part II Solns
6. (4.1 #5abcd)
(a) U is not closed under scalar multiplication (e.g., [1, 1, 1]T is in U , but 2[1, 1, 1]T = [2, 2, 2]T is not
in U ), so U is not a subspace.
(b) U is the set of multiples of [0, 1, 0]T , so U is the span of t
Problem Set 8 Part II Solutions
7. (4.2 #7abc)
(a) False. The set of vectors cfw_X, Y, X + Y is dependent, since the third is a linear combination of the
first two.
(b) True. Suppose the set cfw_X, Y, Z is independent. If aX + bY = 0, then aX + bY + 0Z =
Problem Set 6 Part I Solns
1. (4.1 #6a) If X = aY + bZ, then [2
1
0
1]T = a[1
0
0
1]T + b[0
1
0
1]T , so
a + 0b = 2
0a + 1b = 1
0a + 0b = 0
1a + 1b = 1
These equations have the solution a = 2, b = 1, so X = 2Y Z. Thus X is in spancfw_Y, Z.
2. If A is an n
Problem Set 9 Part II Solutions
7. Let W be the set of symmetric 2 2 matrices with real entries.
(a) Show that W is a subspace of M2,2 .
Let A, B W , and let k be a scalar. Then AT = A and B T = B, so (A + B)T = AT + B T = A + B,
so A + B W . Also (kA)T =
Problem Set 8 Part I Solutions
1. (4.2 #9ab)
(a) We look to see if there are a, b, c, not all zero, such that a(X Y ) + b(Y Z) + c(Z X) = 0. This
is the same as (a c)X + (b a)Y + (c b)Z = 0. Since the set cfw_X, Y, Z is independent, this is
equivalent to
Problem Set 9 Part II Solutions
7. (4.5 #11)
(a) If X and Y are orthogonal, then
X Y 2 = (X Y ) (X Y ) = X X Y X X Y + Y Y = X X + Y Y
X + Y 2 = (X Y ) (X Y ) = X X + Y X + X Y + Y Y = X X + Y Y
Hence X Y 2 = X + Y 2 , so X Y = X + Y .
(b) If X Y = X + Y
Problem Set 5 Part II Solutions
7. 3.2 #10a To write v = v1 + v2 , where v1 is parallel to w and v2 is orthogonal to w, we take v1 = projw v
T
T
and v2 = v v1 . Thus for v = ( 1 1 3 ) and w = ( 2 1 1 ) we have
v1 = projw v =
wv
w 2
v2 = v v1 = ( 1
w = 6w
Problem Set 2 Part II Solutions
Part II
8. Prove: If AB is invertible, then A is invertible and B is invertible. (Hint: write down what the denition
of invertibility of AB tells you.)
If AB is invertible there is a matrix C such that C(AB) = I and (AB)C =
24
Isomorphisms of vector spaces
Linear transformations of vector spaces
Definition 24.1. If V, W are vector spaces, and T : V W is a transformation
from V to W, then T is linear if T preserves sums and scalar multiples, i.e., for
all vectors v, v1 , v2 a
Problem Set 9 Part I Solutions
1
1. For the matrix A = 1
2
1
2
3
1
3
4
0
4
4
(a) Find a basis for the row space of A and find the dimension for the row space of A.
Subtracting the first row from the second, then twice
the first row
from the last row, th
End of term problems
1. If X and Y are column vectors, then the dot product is given by X Y = X t Y . (The transpose makes
the column matrix X into a row matrix X t , and then multiplying that row matrix by the column matrix
Y gives a 1 1 matrix X t Y , w
Review Problems for Exam 3
1. Let cfw_~v1 , ~v2 , ~v3 be a basis of a vector space V . Show that cfw_~v1 , ~v1 + ~v2 , ~v1 + ~v2 + ~v3 is
a basis of V .
2. Find a basis for the nullspace and rowspace of the matrix
1 2 3
1 1 0 .
5 1 6
3. If ~u, ~v , w
~
Problem Set 1
All references are to the Nicholson text. Problems from section 1.4, and problem 6 below, involve matrix
multiplication, which we will discuss Monday. Note that if you are using the print on demand copy of the text
there are two sets of page
Problem Set 3
Show all work. Do all work by hand, unless you are instructed to use the computer (i.e., Wolfram Alpha).
Part I
1. 2.1 #17
2. 2.2 #2
3. 2.2 #4acg
4. 2.2 #12a
5. 2.2 #14a (You dont need to find the inverse.)
6. 2.3 #2a
7. 2.3 #5a
8. 2.3 #7bh
Problem Set 2
Instructions Problems with a star must be done solely by yourself, without discussing with other students
or with me.
Part I
1. (1.5 #2d, page 47)
2. (1.5 #7b)
3. If A is invertible, show that AT is invertible, and that (AT )1 = (A1 )T .
1 2
Problem Set 5
Part I
1. (a) Find all vectors X in R4 that are orthogonal to (1, 1, 1, 1) and (1, 2, 3, 4).
(b) Find two nonzero vectors X1 and X2 in R4 such that X1 and X2 are orthogonal to each other, and
are also orthogonal to both (1, 1, 1, 1) and (1,
Problem Set 6
Part I
1. (4.1 #6a)
2. If A is an n n matrix and is an eigenvalue of A, the eigenspace E is the set of vectors X Rn such
that AX = X. (These are the eigenvectors of A associated with the eigenvalue , together with the
zero vector). Show that
Problem Set 4 and Review Problems for Exam 1
These review problems cover many, but not all topics; make sure to also look at quizzes and
problem sets.
Part I (there is no Part II for this problem set)
1. Solve the following set of equations:
3x y + z
x +
Problem Set 10 Part I Solutions
1. (4.5 #7c)
Let w
~ 1 = (1, 1, 1), w
~ 2 = (1, 0, 1), w
~ 3 = (1, 1, 2). We construct an orthogonal basis ~v1 , ~v2 , ~v3 using the
Gram-Schmidt process.
~v1 = w
~ 1 = (1, 1, 1)
~v2 = w
~ 2 P roj~v1 w
~ 2 = (1, 0, 1) P roj
Problem Set 2 Part I Solutions
Part I
1. (1.5 #2d, page 47)
1
To find the inverse of 1
0
3 1
0 0
2 1
Start with the augmented matrix
1
1
0
3 1
0 0
2 1
1 0 0
0 1 0
0 0 1
After appropriate row operations, one finds that the reduced form is
1 0 0 0
0 1 0 1
0
25
Inner products
The following is a generalization of dot products.
Denition 25.1. An inner product < , > on a real vector space V is a rule
which assigns a real number <v, w> to each pair of vectors v, w in V, in such
a way that
(1) <v, w> = <w,v>
(2) <
8
Determinants
Denition of determinants
Denition 8.1. The determinant of a 11 matrix (k) is just k. If A is an nn
matrix, The (ij)-cofactor of an n n matrix A is (1)i+j times the determinant
of the (n 1) (n 1) submatrix found by deleting ith row and jth c
1
Matrix Multiplication
Notation. A vector in Rn is a nite sequence of n real numbers, and is displayed
a1
a2
either as a 1 n column matrix or as an n 1 row matrix. Vectors in R2
an
3
or R will often be denoted by arrows, e.g. i = (1, 0, 0). Vectors
6
Mathematical Induction
The following is a very useful proof technique.
Theorem 6.1. Principle of Mathematical Induction) Suppose S(1), S(2), S(3).
are statements (one for each natural number 1, 2, .), and
(i) S(n0 ) is true (where n0 is a particular int
5
Matrix Inverses
Denition 5.1. A matrix B is the inverse of a matrix A if AB = I = BA.
Example 5.2. Show that A =
3
1
2
1
and B =
1
1
2
3
are inverses.
Theorem 5.3. Let A be a square matrix.
(1). If A has an inverse, A must be square.
(2). Some square ma
17
Denition of a basis of Rn
Denition 17.1. A basis for Rn is a set of vectors v1 , v2 , ., vk which is both
independent and spans Rn .
We have seen before that independence is equivalent to uniqueness of linear
combinations. Thus vectors v1 , v2 , ., vk