Calculating matrix for linear transformation of orthogonal projection onto plane.











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0
down vote

favorite












The question goes like this:



"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane



$$ 2x_1+2x_2+x_3^{}= 0" $$



So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:



$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$



That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$



Basically, what I will do is set up an equation



$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$



So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?



Thanks in advance.






linear-algebra linear-transformations







share|cite|improve this question
























  • Do you mean orthogonal projection?
    – José Carlos Santos
    Nov 21 at 15:30












  • Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
    – Synchrowave
    Nov 21 at 15:32












  • There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
    – José Carlos Santos
    Nov 21 at 15:35










  • I would say that they look for the projection that is parallell to the plane.
    – Synchrowave
    Nov 21 at 15:39










  • So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
    – Synchrowave
    Nov 21 at 15:59















up vote
0
down vote

favorite












The question goes like this:



"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane



$$ 2x_1+2x_2+x_3^{}= 0" $$



So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:



$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$



That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$



Basically, what I will do is set up an equation



$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$



So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?



Thanks in advance.






linear-algebra linear-transformations







share|cite|improve this question
























  • Do you mean orthogonal projection?
    – José Carlos Santos
    Nov 21 at 15:30












  • Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
    – Synchrowave
    Nov 21 at 15:32












  • There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
    – José Carlos Santos
    Nov 21 at 15:35










  • I would say that they look for the projection that is parallell to the plane.
    – Synchrowave
    Nov 21 at 15:39










  • So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
    – Synchrowave
    Nov 21 at 15:59













up vote
0
down vote

favorite









up vote
0
down vote

favorite











The question goes like this:



"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane



$$ 2x_1+2x_2+x_3^{}= 0" $$



So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:



$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$



That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$



Basically, what I will do is set up an equation



$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$



So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?



Thanks in advance.






linear-algebra linear-transformations







share|cite|improve this question















The question goes like this:



"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane



$$ 2x_1+2x_2+x_3^{}= 0" $$



So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:



$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$



That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$



Basically, what I will do is set up an equation



$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$



So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?



Thanks in advance.






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 16:09

























asked Nov 21 at 15:27









Synchrowave

375




375












  • Do you mean orthogonal projection?
    – José Carlos Santos
    Nov 21 at 15:30












  • Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
    – Synchrowave
    Nov 21 at 15:32












  • There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
    – José Carlos Santos
    Nov 21 at 15:35










  • I would say that they look for the projection that is parallell to the plane.
    – Synchrowave
    Nov 21 at 15:39










  • So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
    – Synchrowave
    Nov 21 at 15:59


















  • Do you mean orthogonal projection?
    – José Carlos Santos
    Nov 21 at 15:30












  • Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
    – Synchrowave
    Nov 21 at 15:32












  • There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
    – José Carlos Santos
    Nov 21 at 15:35










  • I would say that they look for the projection that is parallell to the plane.
    – Synchrowave
    Nov 21 at 15:39










  • So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
    – Synchrowave
    Nov 21 at 15:59
















Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30






Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30














Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32






Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32














There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35




There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35












I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39




I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39












So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59




So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59










2 Answers
2






active

oldest

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up vote
1
down vote













Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.



The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.






share|cite|improve this answer





















  • I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
    – Synchrowave
    Nov 22 at 15:28












  • @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
    – amd
    Nov 22 at 19:25


















up vote
0
down vote













Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.



So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that





  1. $P(2,2,1)=(0,0,0)$;


  2. $P(1,0,-2)=(1,0,-2)$;


  3. $P(0,1,-2)=(0,1,-2)$.


A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    up vote
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    down vote













    Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.



    The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.






    share|cite|improve this answer





















    • I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
      – Synchrowave
      Nov 22 at 15:28












    • @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
      – amd
      Nov 22 at 19:25















    up vote
    1
    down vote













    Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.



    The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.






    share|cite|improve this answer





















    • I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
      – Synchrowave
      Nov 22 at 15:28












    • @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
      – amd
      Nov 22 at 19:25













    up vote
    1
    down vote










    up vote
    1
    down vote









    Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.



    The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.






    share|cite|improve this answer












    Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.



    The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 at 21:26









    amd

    29k21050




    29k21050












    • I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
      – Synchrowave
      Nov 22 at 15:28












    • @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
      – amd
      Nov 22 at 19:25


















    • I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
      – Synchrowave
      Nov 22 at 15:28












    • @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
      – amd
      Nov 22 at 19:25
















    I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
    – Synchrowave
    Nov 22 at 15:28






    I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
    – Synchrowave
    Nov 22 at 15:28














    @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
    – amd
    Nov 22 at 19:25




    @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
    – amd
    Nov 22 at 19:25










    up vote
    0
    down vote













    Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.



    So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that





    1. $P(2,2,1)=(0,0,0)$;


    2. $P(1,0,-2)=(1,0,-2)$;


    3. $P(0,1,-2)=(0,1,-2)$.


    A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$






    share|cite|improve this answer



























      up vote
      0
      down vote













      Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.



      So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that





      1. $P(2,2,1)=(0,0,0)$;


      2. $P(1,0,-2)=(1,0,-2)$;


      3. $P(0,1,-2)=(0,1,-2)$.


      A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.



        So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that





        1. $P(2,2,1)=(0,0,0)$;


        2. $P(1,0,-2)=(1,0,-2)$;


        3. $P(0,1,-2)=(0,1,-2)$.


        A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$






        share|cite|improve this answer














        Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.



        So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that





        1. $P(2,2,1)=(0,0,0)$;


        2. $P(1,0,-2)=(1,0,-2)$;


        3. $P(0,1,-2)=(0,1,-2)$.


        A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 16:24

























        answered Nov 21 at 16:11









        José Carlos Santos

        147k22117218




        147k22117218






























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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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