Identity with scalar product

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In one proof of some theorem regarding normal distribution there is used I think an idenitity which states



$$ (ABy,By)=(B^-1ABy,y)$$
where A is symetric and positive, B is orthogonal and y is a vector.



Can someone explain to me why is it so?






linear-algebra







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    up vote
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    down vote

    favorite












    In one proof of some theorem regarding normal distribution there is used I think an idenitity which states



    $$ (ABy,By)=(B^-1ABy,y)$$
    where A is symetric and positive, B is orthogonal and y is a vector.



    Can someone explain to me why is it so?






    linear-algebra







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In one proof of some theorem regarding normal distribution there is used I think an idenitity which states



      $$ (ABy,By)=(B^-1ABy,y)$$
      where A is symetric and positive, B is orthogonal and y is a vector.



      Can someone explain to me why is it so?






      linear-algebra







      share|cite|improve this question













      In one proof of some theorem regarding normal distribution there is used I think an idenitity which states



      $$ (ABy,By)=(B^-1ABy,y)$$
      where A is symetric and positive, B is orthogonal and y is a vector.



      Can someone explain to me why is it so?






      linear-algebra




      linear-algebra






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      share|cite|improve this question











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      asked 1 hour ago









      ryszard eggink

      657




      657




















          3 Answers
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          down vote



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          It follows easily from

          1. $(u,v)=u^tv$, where $u^t$ is the transpose of $u$,

          2. $(CD)^t=D^tC^t$,

          3. $C$ symmetric iff $C=C^t$, and

          4. $C$ orthogonal iff $C^t=C^-1$.






          share|cite|improve this answer




















          • The fact that $A$ is symmetric is not needed.
            – Joppy
            57 mins ago

















          up vote
          1
          down vote













          An equivalent definition of an orthogonal matrix is one for which $(Bv, Bw) = (v, w)$ for all vectors $v$ and $w$. You can check that if $B$ is orthogonal, then $B^-1$ exists and is orthogonal. Can you work it out from there?






          share|cite|improve this answer



























            up vote
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            I assume $(cdot, cdot) $ denotes the scalar product. An orthogonal matrix $B$ preserves the scalar product by definition, i.e.
            $$(Bx,By)=(x,y).$$
            Since $B^-1$ is also orthogonal, then
            $$(ABy, By) = (B^-1ABy,B^-1By)=(B^-1ABy,y).$$






            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








              up vote
              2
              down vote



              accepted










              It follows easily from

              1. $(u,v)=u^tv$, where $u^t$ is the transpose of $u$,

              2. $(CD)^t=D^tC^t$,

              3. $C$ symmetric iff $C=C^t$, and

              4. $C$ orthogonal iff $C^t=C^-1$.






              share|cite|improve this answer




















              • The fact that $A$ is symmetric is not needed.
                – Joppy
                57 mins ago














              up vote
              2
              down vote



              accepted










              It follows easily from

              1. $(u,v)=u^tv$, where $u^t$ is the transpose of $u$,

              2. $(CD)^t=D^tC^t$,

              3. $C$ symmetric iff $C=C^t$, and

              4. $C$ orthogonal iff $C^t=C^-1$.






              share|cite|improve this answer




















              • The fact that $A$ is symmetric is not needed.
                – Joppy
                57 mins ago












              up vote
              2
              down vote



              accepted







              up vote
              2
              down vote



              accepted






              It follows easily from

              1. $(u,v)=u^tv$, where $u^t$ is the transpose of $u$,

              2. $(CD)^t=D^tC^t$,

              3. $C$ symmetric iff $C=C^t$, and

              4. $C$ orthogonal iff $C^t=C^-1$.






              share|cite|improve this answer












              It follows easily from

              1. $(u,v)=u^tv$, where $u^t$ is the transpose of $u$,

              2. $(CD)^t=D^tC^t$,

              3. $C$ symmetric iff $C=C^t$, and

              4. $C$ orthogonal iff $C^t=C^-1$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 1 hour ago









              Gerry Myerson

              144k8145296




              144k8145296











              • The fact that $A$ is symmetric is not needed.
                – Joppy
                57 mins ago
















              • The fact that $A$ is symmetric is not needed.
                – Joppy
                57 mins ago















              The fact that $A$ is symmetric is not needed.
              – Joppy
              57 mins ago




              The fact that $A$ is symmetric is not needed.
              – Joppy
              57 mins ago










              up vote
              1
              down vote













              An equivalent definition of an orthogonal matrix is one for which $(Bv, Bw) = (v, w)$ for all vectors $v$ and $w$. You can check that if $B$ is orthogonal, then $B^-1$ exists and is orthogonal. Can you work it out from there?






              share|cite|improve this answer
























                up vote
                1
                down vote













                An equivalent definition of an orthogonal matrix is one for which $(Bv, Bw) = (v, w)$ for all vectors $v$ and $w$. You can check that if $B$ is orthogonal, then $B^-1$ exists and is orthogonal. Can you work it out from there?






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  An equivalent definition of an orthogonal matrix is one for which $(Bv, Bw) = (v, w)$ for all vectors $v$ and $w$. You can check that if $B$ is orthogonal, then $B^-1$ exists and is orthogonal. Can you work it out from there?






                  share|cite|improve this answer












                  An equivalent definition of an orthogonal matrix is one for which $(Bv, Bw) = (v, w)$ for all vectors $v$ and $w$. You can check that if $B$ is orthogonal, then $B^-1$ exists and is orthogonal. Can you work it out from there?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Joppy

                  4,855420




                  4,855420




















                      up vote
                      1
                      down vote













                      I assume $(cdot, cdot) $ denotes the scalar product. An orthogonal matrix $B$ preserves the scalar product by definition, i.e.
                      $$(Bx,By)=(x,y).$$
                      Since $B^-1$ is also orthogonal, then
                      $$(ABy, By) = (B^-1ABy,B^-1By)=(B^-1ABy,y).$$






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        I assume $(cdot, cdot) $ denotes the scalar product. An orthogonal matrix $B$ preserves the scalar product by definition, i.e.
                        $$(Bx,By)=(x,y).$$
                        Since $B^-1$ is also orthogonal, then
                        $$(ABy, By) = (B^-1ABy,B^-1By)=(B^-1ABy,y).$$






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          I assume $(cdot, cdot) $ denotes the scalar product. An orthogonal matrix $B$ preserves the scalar product by definition, i.e.
                          $$(Bx,By)=(x,y).$$
                          Since $B^-1$ is also orthogonal, then
                          $$(ABy, By) = (B^-1ABy,B^-1By)=(B^-1ABy,y).$$






                          share|cite|improve this answer












                          I assume $(cdot, cdot) $ denotes the scalar product. An orthogonal matrix $B$ preserves the scalar product by definition, i.e.
                          $$(Bx,By)=(x,y).$$
                          Since $B^-1$ is also orthogonal, then
                          $$(ABy, By) = (B^-1ABy,B^-1By)=(B^-1ABy,y).$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Gibbs

                          3,7672624




                          3,7672624



























                               

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