Subject: Some stuff to know for Exam A.
Date: 6 Feb 1998 12:46:54 -0500
From: squash
Hi Folks
Here are some Things to Know:
Greek Alphabet, lower/upper case, and transliteration.
Definition of a group: (V, +, 0).
(ditto for an Abelian group).
Definition of a vectorspace: (V, +, 0, *)
Definitions:
For a subset W, of V, this W is a "subspace" if...
A subset W is a "flat" in V if...
A collection {w_1, ..., w_K} "spans" V if...
A vector b is in Span{w_1, ..., w_K} if ...
Collection {w_1, ..., w_K} is "linearly independent" if...
Computations:
Let u:= (2,3,4)
and v:= (-2, sqrt(3), 5)
and w:= (0 , -5/3, 17)
Is {u, v, w} an independent collection?
Does {u,v,w} span R^3 ?
Give me a vector which is NOT in Span{u,w}.
Here is a matrix: A = [ 2 3 4 5 ; 0 0 -3 3; -2 0 2 0] .
Compute rref(A).
Here is a system of linear equations... Is it consistent?
Here is an equation, Ax=b (where A is a specific matrix, and b is a
specific column vector), whose "unknown" is the column-vector x.
Coordinatize the complete solution set (all the possible values for x)
in this form:
x = a + alpha_1*u_1 + ... + alpha_K * u_K
where a, u_1, ..., u_K are specific vectors (you tell me what they are).
Thus, as I vary the scalars alpha_1, ..., alpha_K, we see ALL solns to
Ax = b
(More to follow)
-J.King
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From squash Sun Feb 8 15:19:15 -0500 1998
Subject: Review of some definitions
Hi Folks
Some folks asked for a review of some of the definitions that I gave in
class. The notions of "span" and of "linear independence" are also in
your text.
Below, I have put in UPPERCASE, words/phrases such as
EACH, EVERY, ALL
SOME, THERE EXISTS, THERE IS
since these quantifiers are sometimes omitted by students, making their
definitions ambiguous.
I will use 00 to denote the 0-vector (the origin) of my vectorspace.
Also, I will use underbar to mean subscript; thus "w_2" means "w sub 2".
Fix a vectorspace (V, +, 00, *). The "*" is sv-multiplication and "+"
is vector-addition.
A subset S of V is a "subspace" if (S, +, 00, *) is itself a vectorspace.
This simply means that you can't get out of S by vector operations. In
other words, for EVERY choice of vectors u and w in S, and EACH scalar
alpha,
both u+w and alpha*u are in S.
================================================================
The "line" through two points u and w in V is the set line(u,w) of points
(1-t)*u + t*w
as t ranges over ALL the reals. (So t is a scalar.) This line is
degenerate if u=w; for then
(1-t)*u + t*w
is simply that "stand still" parameterization of the single point u=w.
================================================================
A subset S of V is a "flat" if
[1] S is non-empty, AND
[2] for EVERY pair u,w in S, the entire line
line(u,w) is in S.
Equivalent to [2]: For EACH u and w in S, and for EACH two scalars alpha
and beta with
alpha + beta = 1,
we have that the linear combination
alpha*u + beta*w is in S.
================================================================
Given a set B of vectors from V, the "span" of B, written Span(B), is
the set of ALL finite linearly combinations of members of B. In other
words, a vector u is in Span(B) if and only if THERE EXIST vectors
w_1,..., w_K in B
and scalars alpha_1, ..., alpha_K, so that the linear combination
(alpha_1 * w_1) + (alpha_2 * w_2) + ... + (alpha_K * w_K)
equals u.
================================================================
A set B of vectors is "linearly independent" if, for EACH choice of vectors
w_1,..., w_K from B, the ONLY solution to equation
(alpha_1 * w_1) + (alpha_2 * w_2) + ... + (alpha_K * w_K) = 00
is the trivial solution
alpha_1 =0, alpha_2 =0, ..., alpha__K =0 .
================================================================
Remember also to know the Greek alphabet.
May all your functions be linear,
-J.King
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From squash Sun Feb 8 15:32:25 -0500 1998
Subject: Exam
Hi Folks
Bring to the exam, blank paper to write you solutions. I'll bring a
stapler from my office.
Also, as far as I am concerned, you are welcome to bring a thermos of
coffee, your lucky pencil, a pillow, or most anything that makes you
comfortable in an exam. Needless to say (but I'm going to say it anyway)
do not bring a calculator, nor a walkman.
Note: I run my exam on an honor code, and I won't be in the room much
of the time. However, I will stop by occasionally for questions.
Tomorrow's exam I don't expect you'll find difficult. Nonetheless,
let me mention the following:
I believe that an exam's most important role (more important than
simply producing a "grade") is as a learning tool.
It is absolutely ok to ask me for help during the exam. It is ok to ask
me if an answer is right (to a computational problem, I probably won't
know; to a conceptual problem, I may answer your question by asking you a
question).
May all your vectorspaces be finite-dimensional,
-J.King