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Math Help - [SOLVED] Proving k*0=0 using linear algebra

  1. #1
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    [SOLVED] Proving k*0=0 using linear algebra

    Prove that in any vector space k 0 = 0, where k is any real number and 0 is the zero vector.

    Do I use the distributivity of scalar multiplication with respect to field addition ??

    (a+b)v= av+bv

    let a=b=0

    (0+0)v=0v+oV
    ?????

    little confused since it's pretty much obvious!

    Thanks
    Last edited by Khonics89; March 15th 2010 at 05:19 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Khonics89 View Post
    Prove that in any vector space k 0 = 0, where k is any real number and 0 is the zero vector.

    I know I have to use the vector axioms..

    AxiomSignificationAssociativity of additionu + (v + w) = (u + v) + w.Commutativity of additionv + w = w + v.Identity element of additionThere exists an element 0V, called the zero vector, such that v + 0 = v for all vV.Inverse elements of additionFor all v ∈ V, there exists an element wV, called the additive inverse of v, such that v + w = 0. The additive inverse is denoted −v.Distributivity of scalar multiplication with respect to vector addition  a(v + w) = av + aw.Distributivity of scalar multiplication with respect to field addition(a + b)v = av + bv.Compatibility of scalar multiplication with field multiplicationa(bv) = (ab)v [nb 3]Identity element of scalar multiplication1v = v, where 1 denotes the multiplicative identity in F.

    Do I use the distributivity of scalar multiplication with respect to field addition ??

    (a+b)v= av+bv

    let a=b=0

    (0+0)v=0v+oV
    ?????

    little confused since it's pretty much obvious!

    Thanks
    \alpha\bold{0}=\alpha(\bold{0}+\bold{0})=\alpha\bo  ld{0}+\alpha\bold{0}\implies \alpha\bold{0}=\bold{0}
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  3. #3
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    Quote Originally Posted by Khonics89 View Post
    Prove that in any vector space k 0 = 0, where k is any real number and 0 is the zero vector.

    I know I have to use the vector axioms..

    AxiomSignificationAssociativity of additionu + (v + w) = (u + v) + w.Commutativity of additionv + w = w + v.Identity element of additionThere exists an element 0V, called the zero vector, such that v + 0 = v for all vV.Inverse elements of additionFor all v ∈ V, there exists an element wV, called the additive inverse of v, such that v + w = 0. The additive inverse is denoted −v.Distributivity of scalar multiplication with respect to vector addition  a(v + w) = av + aw.Distributivity of scalar multiplication with respect to field addition(a + b)v = av + bv.Compatibility of scalar multiplication with field multiplicationa(bv) = (ab)v [nb 3]Identity element of scalar multiplication1v = v, where 1 denotes the multiplicative identity in F.

    Do I use the distributivity of scalar multiplication with respect to field addition ??

    (a+b)v= av+bv

    let a=b=0

    (0+0)v=0v+oV
    ?????

    little confused since it's pretty much obvious!

    Thanks
    hate these little proofs! always make an error
    I would say

    V+0=V
    k.(V+0)=k.V
    Kv+K0=KV

    but kV is a vector in so additive inverse vector exists ,-kV
    so

    -kV+kV+k0=-KV+kV
    0+k0=0
    kO=0
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