Find all invariant lines of an affine transformation in plane












0














Suppose we have affine transformation:



$$begin{align}x^* &= 2x+y-2\
y^* &= -3x- yend{align}$$



How can we find the equations of all invariant lines?



Thanks in advance






linear-transformations affine-geometry







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    0














    Suppose we have affine transformation:



    $$begin{align}x^* &= 2x+y-2\
    y^* &= -3x- yend{align}$$



    How can we find the equations of all invariant lines?



    Thanks in advance






    linear-transformations affine-geometry







    share|cite|improve this question



























      0












      0








      0







      Suppose we have affine transformation:



      $$begin{align}x^* &= 2x+y-2\
      y^* &= -3x- yend{align}$$



      How can we find the equations of all invariant lines?



      Thanks in advance






      linear-transformations affine-geometry







      share|cite|improve this question















      Suppose we have affine transformation:



      $$begin{align}x^* &= 2x+y-2\
      y^* &= -3x- yend{align}$$



      How can we find the equations of all invariant lines?



      Thanks in advance






      linear-transformations affine-geometry




      linear-transformations affine-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 at 16:24









      Andrei

      11.1k21025




      11.1k21025










      asked Nov 24 at 16:19









      ziadxkabakibi

      11




      11






















          3 Answers
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          0














          The equation of a line originally is $y=ax+b$. After transformation is $y^*=a^*x^*+b^*$. Plug in your equations for $x^*$ and $y^*$ into this equation, group the terms so you can rewrite it as $$y=f(a^*,b^*)x+g(a^*,b^*)$$
          Since the line is invariant, you have the same coefficients so $$f(a,b)=a\g(a,b)=b$$






          share|cite|improve this answer





















          • You’re omitting vertical lines ($x=k$).
            – amd
            Nov 24 at 22:20












          • Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
            – Andrei
            Nov 24 at 22:37










          • In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
            – amd
            Nov 24 at 22:39












          • I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
            – Andrei
            Nov 24 at 22:41










          • Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
            – amd
            Nov 24 at 22:46



















          0














          Calling $p = (x,y),, b = (-2,0)$ we have the generic line as



          $$
          p = p_0 + lambda vec v
          $$



          then



          $$
          p_0 + lambda vec v = A( p_0+lambdavec v) + b
          $$



          or



          $$
          A p_0-p_0+b = lambda(vec v-Avec v),,,forall lambdain mathbb{R}
          $$



          so we have



          $$
          (A-I_2)vec v = 0\
          (A-I_2)p_0 + b = 0
          $$



          now solving for $vec v, p_0$ we have the invariant lines.






          share|cite|improve this answer































            0














            Working in homogeneous coordinates, the point transformation $mathbf x'=Mmathbf x$ transforms lines as $mathbf l'=M^{-T}mathbf l$. The invariant lines $mathbf l=(lambda,mu,tau)^T$ are those for which $M^{-T}mathbf l$ is a nonzero multiple of $mathbf l$, so finding them amounts to finding eigenvectors of $M^{-T}$ that correspond to real nonzero eigenvalues. Alternatively, since the homogeneous coordinates are three-dimensional, the invariant lines are the solutions of $mathbf ltimes M^{-T}mathbf l=0$.



            Note that the line at infinity is always an invariant line of an affine transformation: the last column of $M^{-T}$ for such a transformation is $(0,0,1)^T$.






            share|cite|improve this answer























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              3 Answers
              3






              active

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              3 Answers
              3






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              0














              The equation of a line originally is $y=ax+b$. After transformation is $y^*=a^*x^*+b^*$. Plug in your equations for $x^*$ and $y^*$ into this equation, group the terms so you can rewrite it as $$y=f(a^*,b^*)x+g(a^*,b^*)$$
              Since the line is invariant, you have the same coefficients so $$f(a,b)=a\g(a,b)=b$$






              share|cite|improve this answer





















              • You’re omitting vertical lines ($x=k$).
                – amd
                Nov 24 at 22:20












              • Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
                – Andrei
                Nov 24 at 22:37










              • In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
                – amd
                Nov 24 at 22:39












              • I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
                – Andrei
                Nov 24 at 22:41










              • Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
                – amd
                Nov 24 at 22:46
















              0














              The equation of a line originally is $y=ax+b$. After transformation is $y^*=a^*x^*+b^*$. Plug in your equations for $x^*$ and $y^*$ into this equation, group the terms so you can rewrite it as $$y=f(a^*,b^*)x+g(a^*,b^*)$$
              Since the line is invariant, you have the same coefficients so $$f(a,b)=a\g(a,b)=b$$






              share|cite|improve this answer





















              • You’re omitting vertical lines ($x=k$).
                – amd
                Nov 24 at 22:20












              • Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
                – Andrei
                Nov 24 at 22:37










              • In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
                – amd
                Nov 24 at 22:39












              • I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
                – Andrei
                Nov 24 at 22:41










              • Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
                – amd
                Nov 24 at 22:46














              0












              0








              0






              The equation of a line originally is $y=ax+b$. After transformation is $y^*=a^*x^*+b^*$. Plug in your equations for $x^*$ and $y^*$ into this equation, group the terms so you can rewrite it as $$y=f(a^*,b^*)x+g(a^*,b^*)$$
              Since the line is invariant, you have the same coefficients so $$f(a,b)=a\g(a,b)=b$$






              share|cite|improve this answer












              The equation of a line originally is $y=ax+b$. After transformation is $y^*=a^*x^*+b^*$. Plug in your equations for $x^*$ and $y^*$ into this equation, group the terms so you can rewrite it as $$y=f(a^*,b^*)x+g(a^*,b^*)$$
              Since the line is invariant, you have the same coefficients so $$f(a,b)=a\g(a,b)=b$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 24 at 16:30









              Andrei

              11.1k21025




              11.1k21025












              • You’re omitting vertical lines ($x=k$).
                – amd
                Nov 24 at 22:20












              • Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
                – Andrei
                Nov 24 at 22:37










              • In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
                – amd
                Nov 24 at 22:39












              • I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
                – Andrei
                Nov 24 at 22:41










              • Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
                – amd
                Nov 24 at 22:46


















              • You’re omitting vertical lines ($x=k$).
                – amd
                Nov 24 at 22:20












              • Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
                – Andrei
                Nov 24 at 22:37










              • In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
                – amd
                Nov 24 at 22:39












              • I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
                – Andrei
                Nov 24 at 22:41










              • Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
                – amd
                Nov 24 at 22:46
















              You’re omitting vertical lines ($x=k$).
              – amd
              Nov 24 at 22:20






              You’re omitting vertical lines ($x=k$).
              – amd
              Nov 24 at 22:20














              Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
              – Andrei
              Nov 24 at 22:37




              Yes. I should have mentioned that you need to check that. It is trivial to show that it is not invariant
              – Andrei
              Nov 24 at 22:37












              In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
              – amd
              Nov 24 at 22:39






              In this case, yes, but you’re presenting $y=ax+b$ as the generic equation of any line whatsoever.
              – amd
              Nov 24 at 22:39














              I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
              – Andrei
              Nov 24 at 22:41




              I could have done $ax+by+c=0$, but then I need to check the ratio of the coefficients, since $2ax+2by+2c=0$ is the same line.
              – Andrei
              Nov 24 at 22:41












              Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
              – amd
              Nov 24 at 22:46




              Either way you have to make an extra check beyond what you’ve got here: on the one hand you have to deal separately with vertical lines, on the other you have to examine ratios.
              – amd
              Nov 24 at 22:46











              0














              Calling $p = (x,y),, b = (-2,0)$ we have the generic line as



              $$
              p = p_0 + lambda vec v
              $$



              then



              $$
              p_0 + lambda vec v = A( p_0+lambdavec v) + b
              $$



              or



              $$
              A p_0-p_0+b = lambda(vec v-Avec v),,,forall lambdain mathbb{R}
              $$



              so we have



              $$
              (A-I_2)vec v = 0\
              (A-I_2)p_0 + b = 0
              $$



              now solving for $vec v, p_0$ we have the invariant lines.






              share|cite|improve this answer




























                0














                Calling $p = (x,y),, b = (-2,0)$ we have the generic line as



                $$
                p = p_0 + lambda vec v
                $$



                then



                $$
                p_0 + lambda vec v = A( p_0+lambdavec v) + b
                $$



                or



                $$
                A p_0-p_0+b = lambda(vec v-Avec v),,,forall lambdain mathbb{R}
                $$



                so we have



                $$
                (A-I_2)vec v = 0\
                (A-I_2)p_0 + b = 0
                $$



                now solving for $vec v, p_0$ we have the invariant lines.






                share|cite|improve this answer


























                  0












                  0








                  0






                  Calling $p = (x,y),, b = (-2,0)$ we have the generic line as



                  $$
                  p = p_0 + lambda vec v
                  $$



                  then



                  $$
                  p_0 + lambda vec v = A( p_0+lambdavec v) + b
                  $$



                  or



                  $$
                  A p_0-p_0+b = lambda(vec v-Avec v),,,forall lambdain mathbb{R}
                  $$



                  so we have



                  $$
                  (A-I_2)vec v = 0\
                  (A-I_2)p_0 + b = 0
                  $$



                  now solving for $vec v, p_0$ we have the invariant lines.






                  share|cite|improve this answer














                  Calling $p = (x,y),, b = (-2,0)$ we have the generic line as



                  $$
                  p = p_0 + lambda vec v
                  $$



                  then



                  $$
                  p_0 + lambda vec v = A( p_0+lambdavec v) + b
                  $$



                  or



                  $$
                  A p_0-p_0+b = lambda(vec v-Avec v),,,forall lambdain mathbb{R}
                  $$



                  so we have



                  $$
                  (A-I_2)vec v = 0\
                  (A-I_2)p_0 + b = 0
                  $$



                  now solving for $vec v, p_0$ we have the invariant lines.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 at 18:43

























                  answered Nov 24 at 18:35









                  Cesareo

                  8,1983516




                  8,1983516























                      0














                      Working in homogeneous coordinates, the point transformation $mathbf x'=Mmathbf x$ transforms lines as $mathbf l'=M^{-T}mathbf l$. The invariant lines $mathbf l=(lambda,mu,tau)^T$ are those for which $M^{-T}mathbf l$ is a nonzero multiple of $mathbf l$, so finding them amounts to finding eigenvectors of $M^{-T}$ that correspond to real nonzero eigenvalues. Alternatively, since the homogeneous coordinates are three-dimensional, the invariant lines are the solutions of $mathbf ltimes M^{-T}mathbf l=0$.



                      Note that the line at infinity is always an invariant line of an affine transformation: the last column of $M^{-T}$ for such a transformation is $(0,0,1)^T$.






                      share|cite|improve this answer




























                        0














                        Working in homogeneous coordinates, the point transformation $mathbf x'=Mmathbf x$ transforms lines as $mathbf l'=M^{-T}mathbf l$. The invariant lines $mathbf l=(lambda,mu,tau)^T$ are those for which $M^{-T}mathbf l$ is a nonzero multiple of $mathbf l$, so finding them amounts to finding eigenvectors of $M^{-T}$ that correspond to real nonzero eigenvalues. Alternatively, since the homogeneous coordinates are three-dimensional, the invariant lines are the solutions of $mathbf ltimes M^{-T}mathbf l=0$.



                        Note that the line at infinity is always an invariant line of an affine transformation: the last column of $M^{-T}$ for such a transformation is $(0,0,1)^T$.






                        share|cite|improve this answer


























                          0












                          0








                          0






                          Working in homogeneous coordinates, the point transformation $mathbf x'=Mmathbf x$ transforms lines as $mathbf l'=M^{-T}mathbf l$. The invariant lines $mathbf l=(lambda,mu,tau)^T$ are those for which $M^{-T}mathbf l$ is a nonzero multiple of $mathbf l$, so finding them amounts to finding eigenvectors of $M^{-T}$ that correspond to real nonzero eigenvalues. Alternatively, since the homogeneous coordinates are three-dimensional, the invariant lines are the solutions of $mathbf ltimes M^{-T}mathbf l=0$.



                          Note that the line at infinity is always an invariant line of an affine transformation: the last column of $M^{-T}$ for such a transformation is $(0,0,1)^T$.






                          share|cite|improve this answer














                          Working in homogeneous coordinates, the point transformation $mathbf x'=Mmathbf x$ transforms lines as $mathbf l'=M^{-T}mathbf l$. The invariant lines $mathbf l=(lambda,mu,tau)^T$ are those for which $M^{-T}mathbf l$ is a nonzero multiple of $mathbf l$, so finding them amounts to finding eigenvectors of $M^{-T}$ that correspond to real nonzero eigenvalues. Alternatively, since the homogeneous coordinates are three-dimensional, the invariant lines are the solutions of $mathbf ltimes M^{-T}mathbf l=0$.



                          Note that the line at infinity is always an invariant line of an affine transformation: the last column of $M^{-T}$ for such a transformation is $(0,0,1)^T$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 24 at 22:59

























                          answered Nov 24 at 22:29









                          amd

                          29.1k21050




                          29.1k21050






























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