Alternate inner products on Euclidean space?












3












$begingroup$


After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).



But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.



Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?






linear-algebra inner-product-space







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  • 1




    $begingroup$
    Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
    $endgroup$
    – Daniel Schepler
    Apr 15 at 17:09


















3












$begingroup$


After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).



But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.



Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?






linear-algebra inner-product-space







share|cite|improve this question









New contributor




rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
    $endgroup$
    – Daniel Schepler
    Apr 15 at 17:09
















3












3








3


1



$begingroup$


After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).



But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.



Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?






linear-algebra inner-product-space







share|cite|improve this question









New contributor




rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).



But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.



Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?






linear-algebra inner-product-space




linear-algebra inner-product-space






share|cite|improve this question









New contributor




rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Apr 15 at 16:34









Björn Friedrich

2,70661831




2,70661831






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asked Apr 15 at 15:44









rampatowlrampatowl

1183




1183




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New contributor





rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
    $endgroup$
    – Daniel Schepler
    Apr 15 at 17:09
















  • 1




    $begingroup$
    Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
    $endgroup$
    – Daniel Schepler
    Apr 15 at 17:09










1




1




$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09






$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09












3 Answers
3






active

oldest

votes


















3












$begingroup$

Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
    $endgroup$
    – eyeballfrog
    Apr 15 at 19:01










  • $begingroup$
    Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
    $endgroup$
    – mihaild
    Apr 15 at 19:44



















2












$begingroup$

There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
    $$
    left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
    $$
    Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.



    In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
        $endgroup$
        – eyeballfrog
        Apr 15 at 19:01










      • $begingroup$
        Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
        $endgroup$
        – mihaild
        Apr 15 at 19:44
















      3












      $begingroup$

      Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
        $endgroup$
        – eyeballfrog
        Apr 15 at 19:01










      • $begingroup$
        Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
        $endgroup$
        – mihaild
        Apr 15 at 19:44














      3












      3








      3





      $begingroup$

      Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.






      share|cite|improve this answer









      $endgroup$



      Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Apr 15 at 16:15









      mihaildmihaild

      1,37413




      1,37413












      • $begingroup$
        No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
        $endgroup$
        – eyeballfrog
        Apr 15 at 19:01










      • $begingroup$
        Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
        $endgroup$
        – mihaild
        Apr 15 at 19:44


















      • $begingroup$
        No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
        $endgroup$
        – eyeballfrog
        Apr 15 at 19:01










      • $begingroup$
        Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
        $endgroup$
        – mihaild
        Apr 15 at 19:44
















      $begingroup$
      No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
      $endgroup$
      – eyeballfrog
      Apr 15 at 19:01




      $begingroup$
      No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
      $endgroup$
      – eyeballfrog
      Apr 15 at 19:01












      $begingroup$
      Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
      $endgroup$
      – mihaild
      Apr 15 at 19:44




      $begingroup$
      Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
      $endgroup$
      – mihaild
      Apr 15 at 19:44











      2












      $begingroup$

      There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.






          share|cite|improve this answer









          $endgroup$



          There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 15 at 16:22









          gandalf61gandalf61

          9,323825




          9,323825























              1












              $begingroup$

              For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
              $$
              left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
              $$
              Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.



              In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
                $$
                left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
                $$
                Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.



                In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
                  $$
                  left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
                  $$
                  Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.



                  In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.






                  share|cite|improve this answer









                  $endgroup$



                  For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
                  $$
                  left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
                  $$
                  Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.



                  In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 15 at 19:14









                  eyeballfrogeyeballfrog

                  7,322633




                  7,322633






















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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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