Vector Spaces

More linear algebra time woohoo!

So,

Vector spaces, to my knowledge, are like the big important things about linear algebra that everyone cares about. How to make them. How to manipulate them. What they can do. So on and so forth.

So, what is a vector space?

We’ll,

For a collection of vectors to be called a vectors space it must meet the 3 criteria

Inclusion of the null space. The collection must contain a vector that looks like <0,0,0>.
Closure under addition. If you take two vectors in the vector space and add them together, their sum must also be in the vector space
Closure under scalar multiplication. If you take a vector in the vector space and multiply it by a scalar [1], like this:
2 • <1,2,3>
Then the product, <2,4,6> must also be in the vector space. [2]

So, armed with the knowledge of what it means to be a vector space, we can try and prove that a collection is a vector space ourselves.

Suppose there is a real set of three-dimensional vectors, S = { : x + y + z = 0}. Based on our definition of the set, we know that all vectors in it are of the arbitrary form because all of its terms must sum to zero to be in the set. With this knowledge we can begin looking at the requirements:

Is the zero vector in the set? Well, the vector <0,0,0> is a three-dimensional vectors where all the terms add to 0 so it must be in the set.
Is the set closed under addition? If we take two arbitrary vectors in the set and and add them together we get the vector sum <(x + a), (y + b), -(x + a) -(y+b> and off we add all the terms of that sum together we get 0, so it looks like adding any two vectors in the set will return a vector in the set and it’s therefore closed under addition
Finally, is the set closed under scalar multiplication? If we take and arbitrary vector in the set, , and an arbitrary real number, a, then multiply a • , we get as the product. Once again, if we add all these terms together we see that they sum to zero, so the product of any real number times any vector in the set is also in the set and the set is closed under scalar multiplication.

Because our set of vectors meet the three requirements, it can be considered a vector space and has a lot of cool properties that we can talk about in another blog. For now hopefully you thought the example of a proof for a vector space was cool, annnd have a good week until my next blog.

[1] – for our purposes you can think of a scalar as any real number, it just basically means a number that isn’t a vector.

[2] – the actual requirements can get a little more abstract, so there are conceivably vector spaces that wouldn’t really fit the stuff I said here but this is a good introductory way to understand vector spaces.