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anonymous
 4 years ago
Determine if V is a vector space when:
x+y=xy
cx=x^c
If not state all the vector axioms it fails
anonymous
 4 years ago
Determine if V is a vector space when: x+y=xy cx=x^c If not state all the vector axioms it fails

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If yes verify each vector space axiom

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3here's at least one that fails\[c(\vec u+\vec v)\neq c\vec u+c\vec v\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok what abt tat scaler multiplication?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0x^c is not a linear operation

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3actually it's consistent though... \[c(\vec u+\vec v)= c\vec u+c\vec v\]because we have\[c(\vec u+\vec v)=(uv)^c=u^cv^c= c\vec u+c\vec v\]so I don't see what's wrong with it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well it is definitely not closed under scaler multilication though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because x^c is exponential

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3but that doesn't necessarily take it out of V

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3*the vector space I mean

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ya? ok I dont get a thing, I feel so stupid

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3aha! it is a vector space I thought I'd seen it befor check it out, the zero vector turns out to be 1 I think http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx

phi
 4 years ago
Best ResponseYou've already chosen the best response.0@Tur how do you get from (uv)^c to cu + cv?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup i see it. Thanks :D

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3@ phi that is what you get when you put the two rules together the exponent is distributed because vector addition is defined as multiplication

phi
 4 years ago
Best ResponseYou've already chosen the best response.0nvm, i'll look at your link...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL i htink i am fundamentally lacking basic knowledge

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3I'm not so good at this stuff, I had it wrong at first as well
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