How to show that the dot product of $h(x)$ and $h(y)$ can be expressed as $h(x)^tr * h(y)$?

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In the book of Analysis on Manifolds by Munkres, at page 175, it is given that




Let $h(x) = A cdot x$, where $A$ is orthogonal n-by-n matrix; we show
$h$ is an isometry by showing that is preserves the dot product.Now
the dot product of $h(x)$ and $h(y)$ can be expressed as



$$h(x)^tr * h(y)$$ [...]




But I do not understand how can
$$h(x) cdot h(y) = h(x)^tr * h(y)$$ ?




real-analysis linear-algebra determinant isometry




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  • Expand the expressions in both members (possibly with a numerical example) and compare.
    – Yves Daoust
    Jul 27 at 12:00















up vote
0
down vote

favorite












In the book of Analysis on Manifolds by Munkres, at page 175, it is given that




Let $h(x) = A cdot x$, where $A$ is orthogonal n-by-n matrix; we show
$h$ is an isometry by showing that is preserves the dot product.Now
the dot product of $h(x)$ and $h(y)$ can be expressed as



$$h(x)^tr * h(y)$$ [...]




But I do not understand how can
$$h(x) cdot h(y) = h(x)^tr * h(y)$$ ?




real-analysis linear-algebra determinant isometry




share|cite|improve this question



















  • Expand the expressions in both members (possibly with a numerical example) and compare.
    – Yves Daoust
    Jul 27 at 12:00













up vote
0
down vote

favorite









up vote
0
down vote

favorite











In the book of Analysis on Manifolds by Munkres, at page 175, it is given that




Let $h(x) = A cdot x$, where $A$ is orthogonal n-by-n matrix; we show
$h$ is an isometry by showing that is preserves the dot product.Now
the dot product of $h(x)$ and $h(y)$ can be expressed as



$$h(x)^tr * h(y)$$ [...]




But I do not understand how can
$$h(x) cdot h(y) = h(x)^tr * h(y)$$ ?




real-analysis linear-algebra determinant isometry




share|cite|improve this question











In the book of Analysis on Manifolds by Munkres, at page 175, it is given that




Let $h(x) = A cdot x$, where $A$ is orthogonal n-by-n matrix; we show
$h$ is an isometry by showing that is preserves the dot product.Now
the dot product of $h(x)$ and $h(y)$ can be expressed as



$$h(x)^tr * h(y)$$ [...]




But I do not understand how can
$$h(x) cdot h(y) = h(x)^tr * h(y)$$ ?





real-analysis linear-algebra determinant isometry





share|cite|improve this question










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asked Jul 27 at 11:44









onurcanbektas

2,9641831




2,9641831











  • Expand the expressions in both members (possibly with a numerical example) and compare.
    – Yves Daoust
    Jul 27 at 12:00

















  • Expand the expressions in both members (possibly with a numerical example) and compare.
    – Yves Daoust
    Jul 27 at 12:00
















Expand the expressions in both members (possibly with a numerical example) and compare.
– Yves Daoust
Jul 27 at 12:00





Expand the expressions in both members (possibly with a numerical example) and compare.
– Yves Daoust
Jul 27 at 12:00











2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










An example, $n=4$. If you have two $4 times 1$ matrices,
$$
mathbfu = beginbmatrix
a_1\a_2\a_3\a_4
endbmatrix
qquad
mathbfv= beginbmatrix
b_1\b_2\b_3\b_4
endbmatrix
$$
Then the transpose of $mathbfu$ is a $1 times 4$ matrix
$$
mathbfu^mathrmtr =
beginbmatrix
a_1quad a_2quad a_3quad a_4
endbmatrix
$$
When you multiply matrices, you get a $1 times 1$ matrix
$$
mathbfu^mathrmtr;mathbfv =
left[a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4right]
$$
Then, by convention, we consider a $1 times 1$ matrix to be the same thing as a scalar, so we have the dot product $mathbfu,cdot,mathbfv$.






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    $h(x),,h(y)$ are represented by $ntimes 1$ matrices, so $;^mathrm th(x)cdot h(x);$ is the product of a $1times n$ matrix by an $ntimes 1$ matrix, i.e. a number.






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      An example, $n=4$. If you have two $4 times 1$ matrices,
      $$
      mathbfu = beginbmatrix
      a_1\a_2\a_3\a_4
      endbmatrix
      qquad
      mathbfv= beginbmatrix
      b_1\b_2\b_3\b_4
      endbmatrix
      $$
      Then the transpose of $mathbfu$ is a $1 times 4$ matrix
      $$
      mathbfu^mathrmtr =
      beginbmatrix
      a_1quad a_2quad a_3quad a_4
      endbmatrix
      $$
      When you multiply matrices, you get a $1 times 1$ matrix
      $$
      mathbfu^mathrmtr;mathbfv =
      left[a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4right]
      $$
      Then, by convention, we consider a $1 times 1$ matrix to be the same thing as a scalar, so we have the dot product $mathbfu,cdot,mathbfv$.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        An example, $n=4$. If you have two $4 times 1$ matrices,
        $$
        mathbfu = beginbmatrix
        a_1\a_2\a_3\a_4
        endbmatrix
        qquad
        mathbfv= beginbmatrix
        b_1\b_2\b_3\b_4
        endbmatrix
        $$
        Then the transpose of $mathbfu$ is a $1 times 4$ matrix
        $$
        mathbfu^mathrmtr =
        beginbmatrix
        a_1quad a_2quad a_3quad a_4
        endbmatrix
        $$
        When you multiply matrices, you get a $1 times 1$ matrix
        $$
        mathbfu^mathrmtr;mathbfv =
        left[a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4right]
        $$
        Then, by convention, we consider a $1 times 1$ matrix to be the same thing as a scalar, so we have the dot product $mathbfu,cdot,mathbfv$.






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          An example, $n=4$. If you have two $4 times 1$ matrices,
          $$
          mathbfu = beginbmatrix
          a_1\a_2\a_3\a_4
          endbmatrix
          qquad
          mathbfv= beginbmatrix
          b_1\b_2\b_3\b_4
          endbmatrix
          $$
          Then the transpose of $mathbfu$ is a $1 times 4$ matrix
          $$
          mathbfu^mathrmtr =
          beginbmatrix
          a_1quad a_2quad a_3quad a_4
          endbmatrix
          $$
          When you multiply matrices, you get a $1 times 1$ matrix
          $$
          mathbfu^mathrmtr;mathbfv =
          left[a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4right]
          $$
          Then, by convention, we consider a $1 times 1$ matrix to be the same thing as a scalar, so we have the dot product $mathbfu,cdot,mathbfv$.






          share|cite|improve this answer













          An example, $n=4$. If you have two $4 times 1$ matrices,
          $$
          mathbfu = beginbmatrix
          a_1\a_2\a_3\a_4
          endbmatrix
          qquad
          mathbfv= beginbmatrix
          b_1\b_2\b_3\b_4
          endbmatrix
          $$
          Then the transpose of $mathbfu$ is a $1 times 4$ matrix
          $$
          mathbfu^mathrmtr =
          beginbmatrix
          a_1quad a_2quad a_3quad a_4
          endbmatrix
          $$
          When you multiply matrices, you get a $1 times 1$ matrix
          $$
          mathbfu^mathrmtr;mathbfv =
          left[a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4right]
          $$
          Then, by convention, we consider a $1 times 1$ matrix to be the same thing as a scalar, so we have the dot product $mathbfu,cdot,mathbfv$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 27 at 11:57









          GEdgar

          58.4k264163




          58.4k264163




















              up vote
              0
              down vote













              $h(x),,h(y)$ are represented by $ntimes 1$ matrices, so $;^mathrm th(x)cdot h(x);$ is the product of a $1times n$ matrix by an $ntimes 1$ matrix, i.e. a number.






              share|cite|improve this answer

























                up vote
                0
                down vote













                $h(x),,h(y)$ are represented by $ntimes 1$ matrices, so $;^mathrm th(x)cdot h(x);$ is the product of a $1times n$ matrix by an $ntimes 1$ matrix, i.e. a number.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  $h(x),,h(y)$ are represented by $ntimes 1$ matrices, so $;^mathrm th(x)cdot h(x);$ is the product of a $1times n$ matrix by an $ntimes 1$ matrix, i.e. a number.






                  share|cite|improve this answer













                  $h(x),,h(y)$ are represented by $ntimes 1$ matrices, so $;^mathrm th(x)cdot h(x);$ is the product of a $1times n$ matrix by an $ntimes 1$ matrix, i.e. a number.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 27 at 11:57









                  Bernard

                  110k635102




                  110k635102






















                       

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