Constructing an orthogonal projection matrix onto a line in $mathbbR^2$

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How would one go about constructing an orthogonal projection matrix onto a line in $mathbbR^2$ that contains the unit vector $(u_1, u_2)$? Is Gram-Schmidt necessary to do this?







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    How would one go about constructing an orthogonal projection matrix onto a line in $mathbbR^2$ that contains the unit vector $(u_1, u_2)$? Is Gram-Schmidt necessary to do this?







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      How would one go about constructing an orthogonal projection matrix onto a line in $mathbbR^2$ that contains the unit vector $(u_1, u_2)$? Is Gram-Schmidt necessary to do this?







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      How would one go about constructing an orthogonal projection matrix onto a line in $mathbbR^2$ that contains the unit vector $(u_1, u_2)$? Is Gram-Schmidt necessary to do this?









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      edited 3 hours ago









      Rodrigo de Azevedo

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      DoofusAnarchy

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          The orthogonal projection $p$ onto the line $operatornameVect(u=(u_1,u_2)^T)$ is defined by: for $x=(x_1,x_2)^TinBbb R^2$,



          $$p(x)=langle x,urangle u =(x_1u_1+x_2u_2)u$$
          Hence the desired matrix is



          $$P=beginpmatrixu_1^2&u_1u_2\
          u_1u_2& u_2^2endpmatrix$$
          where the first column is $p((1,0)^T)$ and the second is $p((0,1)^T)$.






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            1 Answer
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            1 Answer
            1






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            active

            oldest

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            up vote
            0
            down vote



            accepted










            The orthogonal projection $p$ onto the line $operatornameVect(u=(u_1,u_2)^T)$ is defined by: for $x=(x_1,x_2)^TinBbb R^2$,



            $$p(x)=langle x,urangle u =(x_1u_1+x_2u_2)u$$
            Hence the desired matrix is



            $$P=beginpmatrixu_1^2&u_1u_2\
            u_1u_2& u_2^2endpmatrix$$
            where the first column is $p((1,0)^T)$ and the second is $p((0,1)^T)$.






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              The orthogonal projection $p$ onto the line $operatornameVect(u=(u_1,u_2)^T)$ is defined by: for $x=(x_1,x_2)^TinBbb R^2$,



              $$p(x)=langle x,urangle u =(x_1u_1+x_2u_2)u$$
              Hence the desired matrix is



              $$P=beginpmatrixu_1^2&u_1u_2\
              u_1u_2& u_2^2endpmatrix$$
              where the first column is $p((1,0)^T)$ and the second is $p((0,1)^T)$.






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                The orthogonal projection $p$ onto the line $operatornameVect(u=(u_1,u_2)^T)$ is defined by: for $x=(x_1,x_2)^TinBbb R^2$,



                $$p(x)=langle x,urangle u =(x_1u_1+x_2u_2)u$$
                Hence the desired matrix is



                $$P=beginpmatrixu_1^2&u_1u_2\
                u_1u_2& u_2^2endpmatrix$$
                where the first column is $p((1,0)^T)$ and the second is $p((0,1)^T)$.






                share|cite|improve this answer













                The orthogonal projection $p$ onto the line $operatornameVect(u=(u_1,u_2)^T)$ is defined by: for $x=(x_1,x_2)^TinBbb R^2$,



                $$p(x)=langle x,urangle u =(x_1u_1+x_2u_2)u$$
                Hence the desired matrix is



                $$P=beginpmatrixu_1^2&u_1u_2\
                u_1u_2& u_2^2endpmatrix$$
                where the first column is $p((1,0)^T)$ and the second is $p((0,1)^T)$.







                share|cite|improve this answer













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                share|cite|improve this answer











                answered 11 hours ago









                user296113

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