Alternate inner products on Euclidean space?












3












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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?










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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
    11 hours ago


















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?










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
    11 hours ago
















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?










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






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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.











share|cite|improve this question









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edited 11 hours ago









Björn Friedrich

2,70661831




2,70661831






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asked 12 hours ago









rampatowlrampatowl

1162




1162




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New contributor





rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    11 hours ago
















  • 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
    11 hours ago










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
11 hours ago






$begingroup$
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
11 hours ago












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
    9 hours ago










  • $begingroup$
    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
    8 hours ago



















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
        9 hours ago










      • $begingroup$
        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
        8 hours ago
















      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
        9 hours ago










      • $begingroup$
        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
        8 hours ago














      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 12 hours ago









      mihaildmihaild

      1,03211




      1,03211












      • $begingroup$
        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
        9 hours ago










      • $begingroup$
        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
        8 hours ago


















      • $begingroup$
        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
        9 hours ago










      • $begingroup$
        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
        8 hours ago
















      $begingroup$
      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
      9 hours ago




      $begingroup$
      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
      9 hours ago












      $begingroup$
      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
      8 hours ago




      $begingroup$
      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
      8 hours ago











      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 12 hours ago









          gandalf61gandalf61

          9,293825




          9,293825























              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 9 hours ago









                  eyeballfrogeyeballfrog

                  7,222633




                  7,222633






















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