Determining Vector Spaces: R^2 and M_2,2 Homework Solutions

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Homework Statement


Determine if the following sets are vector spaces

Part A) R^2; u+v = (u1 + v1, u2 + v2); cu = (cu1,cu2)

part b) M_2,2

A + B = |(a_11 + b_11) (a_12 + b_12)| & cA = |ca_11 ca_12|
|(a_21 + b_21) (a_22 + b_22)| |ca_21 ca_22|

sorry this is a little crude but those are both suppose to be 2X2 matricies


Homework Equations





The Attempt at a Solution


Part A) Yes, it is a vector space because it follows the addition & multiplication properties of vector spaces

Part B) yes it is a vector space because the set is closed under matrix addition & sclar multiplication


Are these statements true and a sufficient answer
 
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If they both are closed under scalar multiplication and addition, then you can conclude that they might be a subset of a given vector space. To check whether or not they are vector spaces, you have to go through the 10 axioms. Does it sound familiar?
 
Your solution is not sufficient because you need to go through each and every vector space axiom to prove that it is in fact a vector space (and you need to show where only one axiom fails to prove it is not a vector space).

It is easy to see that pointwise addition is a simple procedure, but I think the point of the exercise is to go through the seven vector space axioms - many are trivial but as far as it stands now your answer is not sufficient.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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