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Vector space

  • #1
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decide whether this is a vector space or not
a(x,y,z) = (2ax,2ay,2az)

all the addition axoims hold easily
for the scalar multiplications axioms
for some real scaral a
[tex] a(x,y,z) = (ax,ay,az) \in 2(ax,ay,az) [/tex]
[tex] a(x_{1}+x_{2},y_{1}+y_{2},z_{1}+z_{2}) = a(x_{1},y_{1},a(z_{1}) + a(x_{2},y_{2},a(z_{2}) \in (2ax,2ay,2az) [/tex] this doesnt seem to hold because it would only be possible if the wo vecotrs being added were not distinct.

[tex] (a+b)(x,y,z) = (ax+bx,ay+by,az+bz) \in (2az,2ay,2az) [/tex]

[tex] a(bx,by,bz) = (ab)(x,y,z) [/tex] this would seem t ohold if b was 2... not not anytrhing else? Not sure here?

multiplcation by 1 gives us what we want. so the last axiom holds
i can post hte axioms if u want

my text says it is not a vector space because of the failure of scalara multiplication where i have stated the doubts mysefl. Are those hte valid reasons for that??
 
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Answers and Replies

  • #2
matt grime
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stunner5000pt said:
decide whether this is a vector space or not
a(x,y,z) = (2ax,2ay,2az)

what the hell are x,y,z, or a?
 
  • #3
HallsofIvy
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stunner5000pt said:
decide whether this is a vector space or not
a(x,y,z) = (2ax,2ay,2az)

all the addition axoims hold easily
for the scalar multiplications axioms
for some real scaral a
[tex] a(x,y,z) = (ax,ay,az) \in 2(ax,ay,az) [/tex]
[tex] a(x_{1}+x_{2},y_{1}+y_{2},z_{1}+z_{2}) = a(x_{1},y_{1},a(z_{1}) + a(x_{2},y_{2},a(z_{2}) \in (2ax,2ay,2az) [/tex] this doesnt seem to hold because it would only be possible if the wo vecotrs being added were not distinct.

[tex] (a+b)(x,y,z) = (ax+bx,ay+by,az+bz) \in (2az,2ay,2az) [/tex]

[tex] a(bx,by,bz) = (ab)(x,y,z) [/tex] this would seem t ohold if b was 2... not not anytrhing else? Not sure here?

multiplcation by 1 gives us what we want. so the last axiom holds
i can post hte axioms if u want

my text says it is not a vector space because of the failure of scalara multiplication where i have stated the doubts mysefl. Are those hte valid reasons for that??
a(x,y,z)= (2ax,2ay,2az) doesn't make sense. Are you asked to show that the set of all (x,y,z) with ordinary addition but scalar multiplication defined by a(x,y,z)= (2a,2ay,2az) is a vector space?
 
  • #4
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HallsofIvy said:
a(x,y,z)= (2ax,2ay,2az) doesn't make sense. Are you asked to show that the set of all (x,y,z) with ordinary addition but scalar multiplication defined by a(x,y,z)= (2a,2ay,2az) is a vector space?
that is what i am asked to prove
 
  • #5
matt grime
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Multiplication by must send v to v, ie 1.v=v for all v in the space.
 
  • #6
HallsofIvy
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stunner5000pt said:
that is what i am asked to prove
I don't believe for a moment that your problem says exactly
"decide whether this is a vector space or not a(x,y,z) = (2ax,2ay,2az)"
There isn't even a set that could be a vector space there!
 

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