Math 441
Exam 1
x1
x2
1. (8 pts) a) Given a vector x
x2
x3
find A so that Ax
0
1
1
0 0
0
x2
with A the indicated n
1
n matrix.
xn
xn
1
x1
x2
1
as a linear combination
xn
x3
xn
x1
x3
b) Express the vector
x2
11
x2
xn
x2
0
0
. (Describe the
x1
0
0
We have
Solutions to Exercises
2
Problem Set 1.1, page 8
1 The combinations give (a) a line in R3
2
3
4
5
6
7
8
9
10
11
(b) a plane in R3 (c) all of R3 .
v C w D .2; 3/ and v w D .6; 1/ will be the diagonals of the parallelogram with v
and w as two sides going ou
Solutions to Exercises
z2 z1 D b1
10 z3 z2 D b2
0 z3 D b3
7
z1 D
z2 D
z3 D
b1
b2
b2
b3
b3
b3
D
"
1
0
0
1
1
0
1
1
1
#"
b1
b2
b3
#
D
1
b
11 The forward differences of the squares are .t C 1/2
t 2 D t 2 C 2t C 1 t 2 D 2t C 1.
Differences of the nth power are
Solutions to Exercises
42
23 As in Problem 22: Row space basis .3; 0; 3/; .1; 1; 2/; column space basis .1; 4; 2/,
24
25
26
27
28
29
30
31
32
.2; 5; 7/; the rank of (3 by 2) times (2 by 3) cannot be larger than the rank of either
factor, so rank 2 and the
Math 441
Exam 3
1. (8 pts) Subspaces of R n are generally described either as the column space of some
given matrix, or the null space of some given matrix.
a) If we define a subspace V as the nullspace of the matrix A
1
1
2
1
1
11
2
express V instead as
The four subspaces as orthogonal complements
Calculating basis for an orthogonal complement - How? Calculating orthonormal basis for
an orthogonal complement - How?
Ax b has a solution, b is in the column space of A, if and only if Hb 0 , b is in the
null
Math 441 Exam 4
5
6
3
1. (12 pts) Given A
4
a) Find the eigenvalues and eigenvectors of A
5
6
3
det
4
2
2 0 , 1, 2
1:
A
6
6
3
3
3
6
3
I
,v
6
1
,v
2
1
2:
A
I
1
b) Write A in diagonalized form, A SDS 1 where D is a diagonal matrix. Then find a
formula
Math 441
Exam 4
Eigenvalues/eigenvectors:
1) I give you a matrix and you find the eigenvalues and eigenvectors
2) I tell you an eigenvalue and ask you to find the corresponding eigenvectors (basis of null
space of A I ) Usually there is only one eigenvect
Math 441 Exam 2
1. (2.5 pts each, total 15) Answer the following simple questions, and provide a brief
explanation.
a) If the vectors in the set S v 1 , . . , v k are linearly independent, what is a basis for their
span, span S ? What is the dimension of
MATH 441.001
Instr. K. Ciesielski
Spring 2011
NAME (print):
FINAL TEST Review
Final Test will start with: Solve the following exercises. Show your work. (No credit will
be given for an answer with no supporting work shown.)
Remember, That Final Test is co
1. a) Notice that the second vector is three times the first, so all linear combinations of the
two vectors can be expressed as a scalar multiple of the first vector. So the result is the line
t 1, 2, 3
b) The result is a plane that can be described as x,
Sec. 1.2
3,4,5,6,8,9,11,12,16,17,21,23
v 1 3, 4 and w 1 8, 6 give unit vectors in the directions of v, w
5
10
v
w
v w v w 1 3, 4 1 8, 6 48 24 , and
Then cos
5
10
50
25
vw
v
w
1 24
cos
0. 283 79 (in radians)
25
Given w 8, 6 the vectors 4, 3 , 3, 4 , 4,
Math 441
Review topics for first exam
Vectors, linear combinations of vectors
Ax as a linear combination of the columns of A, as the dot product of x with the rows of A
Solve Ax b
Express b as a linear combination of vectors (the columns of A )
Matrix mul
General ideas:
Linear independence, span, basis dimension
Linear independence: c 1 v 1 . . . c k v k 0 only when c 1 0 , . . . , c k 0
Equivalent statements (if v s are column vectors):
Ax 0 only if x 0 (A contains v s as columns)
rankA k
Span: (two usage
Solutions to Exercises
27
Problem Set 3.1, page 127
1 x C y y C x and x C .y C z/ .x C y / C z and .c1 C c2 /x c1 x C c2 x .
2 When c.x1 ; x2 / D .cx1 ; 0/, the only broken rule is 1 times x equals x . Rules (1)-(4)
3
4
5
6
7
8
9
10
11
for addition x C y