MATH 240, Section 2, Fall 2002 : Review for Test II, Tuesday, 10/22/2002
Chapter Two, Sections 2.3 - 2.8
You need to know the definitions and theorems,
with these exceptions
(about isomorphisms):
Definition 2.13,
Theorems 2.13, 2.14, 2.15,
Corollary 2.6.
Procedure 1.
(page 116).
To test whether the vectors
v1,
v2, ...,
vk
are linearly independent or linearly dependent:
- Solve the equation
x1 v1 +
x2 v2 + ... +
xk vk = 0.
Note: the vectors end up as columns in a matrix.
If the only solution is all zeros,
then the vectors are linearly independent.
If there is a nonzero solution,
then the vectors are linearly dependent.
Procedure 2.
(page 114).
To check that the vectors
v1,
v2, ...,
vk
span the subspace W:
- Show that for every vector b in W there
is a soluton to
x1v1 +
x2v2 + ... +
xkvk = b.
Procedure 3.
(page 131).
To find a basis for the subspace
span {
v1,
v2, ...,
vk }
by deleting vectors:
- Construct the matrix whose columns
are the coordinate vectors for the
v's
- Row reduce
- Keep the vectors whose column contains a leading 1
The advantage of this procedure is that the answer
consists of some of the vectors in the original set.
Procedure 4.
(page 156).
To find the transition matrix PS<-T:
- Construct the matrix A whose columns
are the coordinate vectors for the basis S
- Construct the matrix B whose columns
are the coordinate vectors for the basis T
- Row reduce the matrix [ A | B ] to the form [ I | P ]
- The matrix P is the transition matrix
The purpose of the procedure is to allow a change of coordinates
[ v ]S
= PS<-T
[ v ]T .
Procedure 5.
(page 143).
To find a basis for the solution space of the system
A x = 0 :
- Row reduce A
- Identify the independent variables in the solution
- In turn, let one of these variables be 1, and all others be 0
- The corresponding solution vectors form a basis
Procedure 6.
To find a simplified basis for the subspace
span {
v1,
v2, ...,
vk } :
- Construct the matrix whose rows are the coordinate vectors for the
v's
- Row reduce
- The nonzero rows form a basis
The advantage of this procedure is that the vectors
in the basis have lots of zeros, so they are in a useful form.
Chapter Three, Sections 3.3, 3.4
You need to know the definitions and theorems,
with these exceptions:
Theorem 3.2;
Theorem 3.8.
Procedure 1. (the Gram-Schmidt Process, page 215).
Let find an orthonormal basis for an m-dimensional subspace of an inner product space V:
Given a basis
S = {
u1,
u2,
. . .
um } for W,
first find an orthogonal basis
T* = {
v1,
v2,
. . .
vm } as follows:
- v1 = u1
- v2 = u2
- [(u2 , v1) /
(v1 , v1) ] v1
- v3 = u3
- [(u3 , v1) /
(v1 , v1) ] v1
- [(u3 , v2) /
(v2 , v2) ] v2
- etc
Finally,
T = {
w1,
w2,
. . .
wm }
where wi is found by dividing vi its length