Analogy of covectors

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Given that the bases of a vector and their duals satisfy:
$$epsilon^ie_j=delta^i_j$$
in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
$$V^*=varphi:V longrightarrow mathbbR$$
with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.



Thanks in advance!



P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces







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    Given that the bases of a vector and their duals satisfy:
    $$epsilon^ie_j=delta^i_j$$
    in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
    $$V^*=varphi:V longrightarrow mathbbR$$
    with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.



    Thanks in advance!



    P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Given that the bases of a vector and their duals satisfy:
      $$epsilon^ie_j=delta^i_j$$
      in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
      $$V^*=varphi:V longrightarrow mathbbR$$
      with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.



      Thanks in advance!



      P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces







      share|cite|improve this question














      Given that the bases of a vector and their duals satisfy:
      $$epsilon^ie_j=delta^i_j$$
      in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
      $$V^*=varphi:V longrightarrow mathbbR$$
      with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.



      Thanks in advance!



      P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces









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      edited Aug 21 at 16:39

























      asked Apr 3 at 17:02









      user518704

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          It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.






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            1 Answer
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            1 Answer
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            up vote
            2
            down vote



            accepted










            It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.






                share|cite|improve this answer












                It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 3 at 17:40









                57Jimmy

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