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Math Help - Bilinear Transformations

  1. #1
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    Bilinear Transformations

    Let f : F^{nxn} x F^{nxn} -> F as follows : F(A,B)=tr(AB).

    Prove that there is no matrix A (other than the 0 matrix) for which  f(A,B)=0 for every B.
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  2. #2
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    Quote Originally Posted by adam63 View Post
    Let f : F^{nxn} x F^{nxn} -> F as follows : f(A,B)=tr(AB).

    Prove that there is no matrix A (other than the 0 matrix) for which  f(A,B)=0 for every B.
    let A=[a_{ij}] and suppose that \{e_{ij} : \ 1 \leq i,j \leq n \} is the standard basis for the vector space of all n \times n matrices over F. now \text{tr}(AB)=0, for all B, if and only if \text{tr}(Ae_{ij})=0, for all

    1 \leq i,j \leq n. but Ae_{ij}=\sum_{k=1}^n a_{ki}e_{kj} and thus \text{tr}(Ae_{ij})=a_{ji}. therefore \text{tr}(Ae_{ij})=0, for all 1 \leq i,j \leq n, if and only if A=0.
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    Well, thank you very much!

    BTW, are you sure there is no other way, more 'elegant'?
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    Quote Originally Posted by NonCommAlg View Post
    let A=[a_{ij}] and suppose that \{e_{ij} : \ 1 \leq i,j \leq n \} is the standard basis for the vector space of all n \times n matrices over F. now \text{tr}(AB)=0, for all B, if and only if \text{tr}(Ae_{ij})=0, for all

    1 \leq i,j \leq n. but Ae_{ij}=\sum_{k=1}^n a_{ki}e_{kj} and thus \text{tr}(Ae_{ij})=a_{ji}. therefore \text{tr}(Ae_{ij})=0, for all 1 \leq i,j \leq n, if and only if A=0.
    Quote Originally Posted by adam63 View Post
    Well, thank you very much!

    BTW, are you sure there is no other way, more 'elegant'?
    Now, I am not the most knowledgeable linear algebraist (? :S) in the world, but I now an "elegant" proof when I see one. That was an elegant proof.
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    Quote Originally Posted by Drexel28 View Post
    Now, I am not the most knowledgeable linear algebraist (? :S) in the world, but I now an "elegant" proof when I see one. That was an elegant proof.
    Thanks !

    Just thought there would be something more 'verbal', or an assumption that would lead to a contradiction.
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    Quote Originally Posted by adam63 View Post
    Well, thank you very much!

    BTW, are you sure there is no other way, more 'elegant'?
    it's math, there's always another way! besides, looking for (possible) more elegant proofs is tempting when the problem is not straightforward.
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