What was the motivation to define what a vector space is?

Tosh5457
Messages
130
Reaction score
28
The book I use for linear algebra explains that the motivation for defining a vector space has to do with the Gauss' reduction method taking linear combinations of the rows, but I don't understand the explanation very well. Can somebody explain?
 
Physics news on Phys.org
Hi Tosh5457! :smile:

The idea behind linear algebra is to provide a theory to solve (or at least to handle) linear equations. Solving linear equation in the real numbers is quite easy using Gauss' method.

Now, a vector space is the most general possible space in which linear equations still make sense and in which they are solvable.

For example, if V is an abstract vector space, then it makes sense to ask which vectors v and w satisfy

\left\{\begin{array}{c} 2v + 3w=0\\ 3v+4w=0\\ \end{array}\right.

This makes sense. Indeed, 2v and 3w makes sense because it is scalar multiplication (and we always have scalar multiplication on a vector space). Also 2v+3w makes sense because it's an addition (and we always have an addition on a vector space). And 0 makes sense since the vector space always has a zero. So this system of equations makes sense.

Furthermore, we can solve this system of equations by exactly the same methods as we would solve a system over \mathbb{R}. Try to solve this system over this general vector space!

So a vector space is a structure in which it is possible to describe linear equations and in which it is possible to solve them. Also, Gauss elimination works in any vector space.

So if we want to study how to solve linear equations, we might as well study them over arbitrary vector spaces. And this is what we do.
 
Adding to what micromass has said, a lot of phenomena act like vectors and when you have the ability to define things in term of an "inner product space", you end up getting a "geometric picture" of how particular objects behave with respect to particular basis.

Fourier series decomposition for periodic functions is a good demonstration of this. We can treat different frequency components as being independent contributions to the total signal so that no other frequency would affect a particular signal in the same way that changing the x co-ordinate of a point doesn't change the y or z co-ordinate of a point.

So with the inner product spaces (which are vector spaces with an inner product) we can actually find out orthonormal basis, and if the inner product is valid, we can treat these objects like arrows and for orthogonal basis, each contribution looks like it is at "right angles" to each other and can pictured like adding up "perpendicular lines" like a right angle triangle.

As a result, this provides a standard way of building decompositions and checking if decompositions are valid by showing orthonormality. Mathematically this is great because it is a way of breaking things down into atoms that are independent, and using this you can do all kinds of things like compressing signals, to doing fast classifications and many other applications.
 
Thanks for the explanations :smile:
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top