The projection of a barycenter onto a plane is the barycenter of the projected triangle?

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When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?



I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.




geometry affine-geometry




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    When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?



    I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.




    geometry affine-geometry




    share|cite|improve this question























      up vote
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      down vote

      favorite









      up vote
      1
      down vote

      favorite











      When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?



      I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.




      geometry affine-geometry




      share|cite|improve this question













      When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?



      I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.





      geometry affine-geometry





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 20 at 19:18









      Arnaud Mortier

      19k22159




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      asked Jul 20 at 13:37









      Alexandre Tourinho

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          The answer is yes, and the reason is both deep and straightforward.



          A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).



          An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.






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            active

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            up vote
            3
            down vote













            The answer is yes, and the reason is both deep and straightforward.



            A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).



            An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.






            share|cite|improve this answer

























              up vote
              3
              down vote













              The answer is yes, and the reason is both deep and straightforward.



              A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).



              An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.






              share|cite|improve this answer























                up vote
                3
                down vote










                up vote
                3
                down vote









                The answer is yes, and the reason is both deep and straightforward.



                A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).



                An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.






                share|cite|improve this answer













                The answer is yes, and the reason is both deep and straightforward.



                A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).



                An easy way to see why such a projection is affine is that if you choose the origin of $Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.







                share|cite|improve this answer













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                share|cite|improve this answer











                answered Jul 20 at 13:48









                Arnaud Mortier

                19k22159




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