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Math Help - Operators on Odd-Dimensional Real Vector Spaces Eigenvalue Existence Proof

  1. #1
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    Operators on Odd-Dimensional Real Vector Spaces Eigenvalue Existence Proof

    I'm having trouble seeing how induction is used in this proof for this theorem which states that every operator on an odd-dimensional real vector space has an eigenvalue.

    The proof starts by supposing V is a real vector space with odd dimension. Then the proof states that it will be using induction in step sizes of 2 on the dimension of V (which I understand since we're proving this for odd dimensions).

    The induction starts with when dimV=1 then the operator on V has an eigenvalue which I understand.

    Then the proof supposes that dimV is an odd number greater than 1. Using induction we can assume that the desired result holds for dimensions 2 less than dimV.

    I can't clearly see how this is true i.e. how, say, if dimV=3 then the operator on V has an eigenvalue.

    Thank you.
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  2. #2
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    Quote Originally Posted by Anthonny View Post
    I'm having trouble seeing how induction is used in this proof for this theorem which states that every operator on an odd-dimensional real vector space has an eigenvalue.


    Hint (the size of an elephant): every real polynomial of odd degree has a real root.

    Oh, and no induction at all needed.

    Tonio



    The proof starts by supposing V is a real vector space with odd dimension. Then the proof states that it will be using induction in step sizes of 2 on the dimension of V (which I understand since we're proving this for odd dimensions).

    The induction starts with when dimV=1 then the operator on V has an eigenvalue which I understand.

    Then the proof supposes that dimV is an odd number greater than 1. Using induction we can assume that the desired result holds for dimensions 2 less than dimV.

    I can't clearly see how this is true i.e. how, say, if dimV=3 then the operator on V has an eigenvalue.

    Thank you.
    .
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  3. #3
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    Quote Originally Posted by tonio View Post
    .
    I haven't really learned eigenvalues using the characteristic polynomial, but rather solving them directly using linear transformations.

    However, I believe from this hint the characteristic polynomial of an operator on an odd-dimensional real vector space is of odd-degree which from this hint implies that it has a real root and therefore an eigenvalue, although I'm not completely sure.

    Although this is a completely valid proof, my text presents the proof using induction

    Thanks anyway
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Anthonny View Post
    Although this is a completely valid proof, my text presents the proof using induction
    Suppose \dim V=2n+1 and consider a basis on V:

    B=\left\{e_1,\ldots,e_{2n-1},e_{2n},e_{2n+1}  \right\}

    Denote: F_1=<e_1,\ldots,e_{2n-1}>,\;F_2=<e_{2n},e_{2n+1}><br />
( so, V=F_1\oplus{F_2} )

    Consider a linear map: T:V\rightarrow{V} . Then:

    \widehat{T}:F_1\rightarrow{F_1},\;\widehat{T}(x)=T  (x) is a linear map.

    Could you continue?.

    Regards.

    Fernando Revilla
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