Navigating Linear Algebra Ep. 2: Abstract Vector Space and Series Overview

We give a rough overview of the series, as well as introducing the notion of the abstract vector space through an axiomatic approach.

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[ummwho]

Hey LSMP, I had a question about Problem 3. I don't think I'm understanding exactly what it says. We consider a "pair" of numbers in R^2 but the field Z/2Z? I have been systematically checking the vector space conditions. I'm going to put what I found below, can you let me know if I'm on the right track?
Checking for the distributive property condition (a+b)v = av + bv for any scalars a, b.
So, in the context of the problem, suppose we pick a and b to be 1 and v to be (1, 1) from R^2.
Then we check that (1+1) (1, 1) = 1(1, 1) + 1(1, 1)

On the left, we have (1+1)(1, 1) = (0)(1, 1) = (0*1, 0*1) = (0, 0) where we have used the fact that in Z/2Z 1+1 = 0 and then did element-wise multiplication and (0, 0) is in R^2
On the right, we have 1*(1, 1) + 1*(1, 1) = (1*1, 1*1) + (1*1, 1*1) = (1, 1) + (1, 1) = (2, 2) which is also in R^2
So therefore the distributive property fails and R^2 is not a vector space over Z/2Z?
Thanks again and sorry for bothering you with all these questions and incorrect answers. This is really fun though, thanks for doing it! I spent all of breakfast thinking that one over! ;)

[martinsanchez-hwfi]

Are there solutions to the proposed problems?

In the first one, it says that C is the vectors set and we choose a field (for example Q) so its a vectorspace C over the field Q. But then it says scalar miltiplication taken as field multiplication in C... How if we have an iperation QxC--> C

[cantcommute]

For problem 4: suppose we have a four element field {0, 1, a, b}. For the sake of contradiction suppose that 1+1 !=0. Then WLOG 1+a=0 and b+b=0 as every inverse is unique. But we must also have from field multiplication that a*b=1 as 0 and 1 cannot be their inverses. But from the distributive property we have a*0=a*(b+b)=a*b+a*b=1+1=0. A contradiction!

[shrayammitra]

I think Z/p is finite iff p is prime because for a modular inverse to exist of any number x, x and p has to be coprime and hence for all x it has to satisfy, so p must be prime, is my explanation correct? Also it's very interesting to think that set of integers can't be a field whereas Z/p is a finite field

[axemenace]

I can't seem to figure out how to construct an infinite field with 1+1=0. I noticed that for any element a in the field, a+a=0, but I cannot proceed. Could you provide a hint?

[ummwho]

Ok, I'm going to edit this comment over the next two days as I answer more and more but I think I got the first part of Problem 1.
Problem 1
Suppose C is a vector space over Q, where the elements of C are complex numbers of the form x + yi where x, y are in Q and i = \sqrt{-1}. Then for the integers a, b, c, d, e, f, g, h we have a/b, c/d, e/f and g/h in Q since as per the video they are of the form a/b where a and b are integers as stated in the video.
Then field addition gives
a/b + (c/d)i + e/f + (g/h)i = (a/b + e/f) + (c/d + g/h)i = ((af + eb) / bf ) + ((ch + gd)/dh)i and since (af + eb)/bf is in Q because (af + eb) is an integer and so is bf, and (ch + gd) is an integer and so is dh, then ((ch + gd)/dh) is also in Q. Therefore field addition is satisfied.

For multiplication, let a, b, c, d, e, f be integers and i same as before. Then for a/b, c/d and e/f in Q,
(a/b)*((c/d) + (e/f)i) = (ac/bd) + (ae/bf)i
Since ac, bd, ae and bf are integers, ac/bd and ae/bf are in Q so multiplication is satisfied.

Good so far? Thanks again LSMP! You're amazing! I'll try problem two in the morning, it's a little late where I am right now. Goodnight and thank you again!

[constexprThoughts]

What is nutritious and commutes? An Abelian soup!

[cantcommute]

I think for problem 3, the trouble is that the axiom (a+b)v=av+bv does not hold if we set a=b=1

(unless every vector is the zero vector lol)

[walkingradiance]

are octonions and hamiltonians a valid field?

[agfd]

I have another fun little problem that you can prove: show that a field cannot have multiple distinct zeros. In other words, if F is a field and 0_1 and 0_2 are zeros, then 0_1=0_2 automatically.