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anilorap

  • 4 years ago

list five elements and determine whether the set is a vector space: (verify the 10 axioms): V={(x,y,z) | z=3}. addition defined by (x1,y1,z1)+(x2,y2,z2) = (x1+x2,y1+y2,3) scalar multiplication defined by s(x,y,z)=(sx,sy,3)

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  1. anonymous
    • 4 years ago
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    list the ten axioms and check. this just fixes z at 3, so it might as well be \[V=\mathbb R^2\]

  2. anilorap
    • 4 years ago
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    yea im in the third axiom.. u+(v+w)

  3. anonymous
    • 4 years ago
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    for example you can check that \[v_1+v_2=v_2+v_1\]

  4. anilorap
    • 4 years ago
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    and i dont know how to add them

  5. anonymous
    • 4 years ago
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    ok \[u+(v+w)=(<x_1,y_1,3>+<x_2,y_2,3>)+<x_3,y_3,3>\]

  6. anonymous
    • 4 years ago
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    addition is given by \[(<x_1,y_1,3>+<x_2,y_2,3>)=<x_1+x_2,y_1+y_2,3>\]

  7. anonymous
    • 4 years ago
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    and then \[(<x_1,y_1,3>+<x_2,y_2,3>)+<x_3,y_3,3>\] \[=<x_1+x_2,y_1+y_2,3>+<x_3,y_3,3>\] \[=<x_1+x_2+x_3,y_1+y_2+y_3,3>\]

  8. anonymous
    • 4 years ago
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    now do it the other way and see that you get the same thing. that is check that \[<x_1,y_1,3>+(<x_2,y_2,3>+<x_3,y_3,3>)=<x_1+x_2+x_3,y_1+y_2+y_3,3>\] as well

  9. anilorap
    • 4 years ago
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    ohh,,, i was adding lol =<x1+x2,y1+y2,3>+<x3,y3,3> =<x1+x2+x3,y1+y2+y3,6>

  10. anilorap
    • 4 years ago
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    stupid me... cool cool i get it

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