# Correct(?) Linear Algebra Proof

1. Jun 8, 2013

### hotcommodity

1. The problem statement, all variables and given/known data

Prove that if a∈F (where F represents ℝ or ℂ), v∈V (where V is a vector space) and av = 0, then a= 0 or v = 0.

2. Relevant equations

The axioms for a vector space may be relevant.

3. The attempt at a solution

Case 1 (v = 0):
Suppose that a∈F, v∈V, and av = 0.
Also, let u∈F.
Then av + au = au (added au to both sides)
So a(v + u) = au (distributive property)
And v + u = u (multiplied both sides by 1/a)
Therefore v = 0.

Case 2 (a = 0):
Suppose that a∈F, v∈V, and av = 0.
I'm stuck here.

The book proves this statement slightly different. For Case 1, the author simply assumes a ≠ 0 and he may therefore multiply both sides by 1/a to get v = 0. For Case 2, the author more or less says av = 0 when a = 0 just because, without any kind of justification.

On a side note, I just finished a proofs book and began self-studying linear algebra for mathematicians. The above problem is from Linear Algebra Done Right (by Axler), which I recently read is targeted at graduate students. As an undergraduate, I'm having a hell of a time so far with this book. If anyone can recommend a more intermediate linear algebra book (undergraduate level, with proofs) I'd surely appreciate it. Thanks in advance.

2. Jun 9, 2013

### haruspex

This is a strange way to start. You are not given that either a=0 or v=0; that's what you're trying to prove. A reasonable start would be "suppose a≠0".
To do that, you have to suppose a≠0.

3. Jun 9, 2013

### lurflurf

^I think that was a mistake, but one might like to show that v=0->av=0 and a=0->av=0 as an exercise and to demonstrate that the truth of the statement is not vacuous (it is not impossible that av=0).

There are a few ways to do this, so think more about it, but one way is..

It may help to first prove that
a0=0
0v=0
hint: Use the distributive property
0v=(0+0)v
a0=a(0+0)
suppose av=0
possibility 1:a=0
possibility 2:a!=0 (!= means not equal to)
there exist b such that ab=ba=1
bav=b0

comment: We have taken advantage of the fact a (when nonzero) has a multiplicative inverse, v does not.

As far as Axler it is far from graduate level. Where did you read that? Update: On the book website it says Content Level » Lower undergraduate This text for a second course in linear algebra is aimed at math majors and graduate students. So it is not graduate level. It is used in freshman honors courses and sophomore/junior courses. It is a fine book, but it has very limited topics, for example F=R or C, spaces are always finite dimensional, multilinear algebra is not covered, Jordan forms are not used for much. A problem with the book is it is aimed at readers at a very particular level. They need to have seen a bit of linear algebra and they need to be developing some feeling for proofs. Readers above or below this level need some outside sources (other books, a teacher, or some notes) to use the book profitably. Here are some other books. I unfortunately do not know of any book that I feel does a good job starting from the very beginning.

Introduction to Linear Algebra (Undergraduate Texts in Mathematics) by Serge Lang
For very basic stuff.
Linear Algebra (Undergraduate Texts in Mathematics) by Serge Lang
A bit higher level close to Axler
Linear Algebra and Its Applications by Gilbert Strang
I do not like this book, others do.
Linear Algebra by Kenneth M Hoffman and Ray Kunze
Well known but dull. Has some good stuff, but does not cover a lot of things.
Finite-Dimensional Vector Spaces by P.R. Halmos
A classic you should look at for sure. Like Axler it is written for a particular level and includes only essential topics.
Linear Algebra by Georgi E. Shilov
A great book. Inexpensive and a good complement to Axler as it includes some easier and harder topics and has a very different approach.
Matrix Analysis by Roger A. Horn and Charles R. Johnson
Introduction to Matrix Analysis (Classics in Applied Mathematics) by Richard Bellman
These two are good books, but a bit more applied and focused on Matrix Analysis.