I was working on two Examples of Friedberg- Insel-Spence's Linear Algebra. In example 6, in RR^2(not 2-dimensional real vector space, consider it as the set RR xx RR) scalar multiplication was defined as usual, but vector addition was defined as the following: (a_1,b_1)+(a_2,b_2)=(a_1+b_2,a_1−b_2) for any (a_1,b_1),(a_2,b_2) in RR^2

Lustyku8

Lustyku8

Answered question

2022-09-24

I was working on two Examples of Friedberg- Insel-Spence's Linear Algebra. In example 6, in R 2 (not 2-dimensional real vector space, consider it as the set R × R ) scalar multiplication was defined as usual, but vector addition was defined as the following:
( a 1 , b 1 ) + ( a 2 , b 2 ) = ( a 1 + b 2 , a 1 b 2 ) for any ( a 1 , b 1 ) , ( a 2 , b 2 ) R 2 . The set R 2 is closed addition and scalar multiplication, but it's not a vector space over R because R 2 is not an abelian group under addition, for instance, this operation is neither commutative nor associative. Moreover, there is an issue with the distribution. Is there any complete list of ways(addition + scalar multiplication) such that the set R 2 is a 2-dimensional vector space?

Answer & Explanation

Malcolm Flores

Malcolm Flores

Beginner2022-09-25Added 8 answers

I think that you are hoping for too much.
Let π : R 2 R 2 be any 1−1 and onto map. (Any map, not necessarily linear or even additive.)
Now define on R 2 a new addition and scalar multiplication as follows:
a b := π 1 ( π ( a ) + π ( b ) ) ,   λ a := π 1 ( λ π ( a ) ) .
Then R 2 is an R vector space of dimension 2 with respect to these operations. You can check the axioms, but as we have just re-labelled all the vectors it is probably "clear" that this is the case.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?