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