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
v_{1},
v_{2}, ...,
v_{k}
are linearly independent or linearly dependent:
- Solve the equation
x_{1} v_{1} +
x_{2} v_{2} + ... +
x_{k} v_{k} = 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
v_{1},
v_{2}, ...,
v_{k}
span the subspace W:
- Show that for every vector b in W there
is a soluton to
x_{1}v_{1} +
x_{2}v_{2} + ... +
x_{k}v_{k} = b.
Procedure 3.
(page 131).
To find a basis for the subspace
span {
v_{1},
v_{2}, ...,
v_{k} }
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 P_{S<-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}
= P_{S<-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 {
v_{1},
v_{2}, ...,
v_{k} } :
- 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 = {
u_{1},
u_{2},
. . .
u_{m} } for W,
first find an orthogonal basis
T* = {
v_{1},
v_{2},
. . .
v_{m} } as follows:
- v_{1} = u_{1}
- v_{2} = u_{2}
- [(u_{2} , v_{1}) /
(v_{1} , v_{1}) ] v_{1}
- v_{3} = u_{3}
- [(u_{3} , v_{1}) /
(v_{1} , v_{1}) ] v_{1}
- [(u_{3} , v_{2}) /
(v_{2} , v_{2}) ] v_{2}
- etc
Finally,
T = {
w_{1},
w_{2},
. . .
w_{m} }
where w_{i} is found by dividing v_{i} its length