Rotation of axes by 45 degrees












4














I was reading a book in which it is mentioned that:



Rotate coordinate axes by $45$ degrees so that a point $(x,y)$ becomes $(x+y,y-x)$ .



Here is image 1



Here is image 2



I can't understand how the new coordinates became $(x+y,y-x)$ .
If I apply the formula for rotation of axes I'll get $sqrt{2}$ $(x-y, x+y) $.










share|cite|improve this question
























  • I agree with you. The distance $BC$ before the rotation is $sqrt {13}$ while it is $sqrt {26}$ afterwards. I think they stretch it back afterwards .
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:21










  • Yes, your are right. But, please explain the context.
    – Martín Vacas Vignolo
    Dec 31 '18 at 6:21










  • @MohammadZuhairKhan I've uploaded the whole section. Can you look something now?
    – Ayaz S Imran
    Dec 31 '18 at 6:25






  • 3




    @MartínVacasVignolo the new coordinates after rotation of the cordinates at an angle f for a point (x,y){x = Xcos(f) - Ysin(f)}{y = Xcos(f) + Ysin(f)}
    – Ayaz S Imran
    Dec 31 '18 at 6:27










  • youtube.com/watch?v=BPgq2AudoEo if you want more details.
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:31
















4














I was reading a book in which it is mentioned that:



Rotate coordinate axes by $45$ degrees so that a point $(x,y)$ becomes $(x+y,y-x)$ .



Here is image 1



Here is image 2



I can't understand how the new coordinates became $(x+y,y-x)$ .
If I apply the formula for rotation of axes I'll get $sqrt{2}$ $(x-y, x+y) $.










share|cite|improve this question
























  • I agree with you. The distance $BC$ before the rotation is $sqrt {13}$ while it is $sqrt {26}$ afterwards. I think they stretch it back afterwards .
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:21










  • Yes, your are right. But, please explain the context.
    – Martín Vacas Vignolo
    Dec 31 '18 at 6:21










  • @MohammadZuhairKhan I've uploaded the whole section. Can you look something now?
    – Ayaz S Imran
    Dec 31 '18 at 6:25






  • 3




    @MartínVacasVignolo the new coordinates after rotation of the cordinates at an angle f for a point (x,y){x = Xcos(f) - Ysin(f)}{y = Xcos(f) + Ysin(f)}
    – Ayaz S Imran
    Dec 31 '18 at 6:27










  • youtube.com/watch?v=BPgq2AudoEo if you want more details.
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:31














4












4








4







I was reading a book in which it is mentioned that:



Rotate coordinate axes by $45$ degrees so that a point $(x,y)$ becomes $(x+y,y-x)$ .



Here is image 1



Here is image 2



I can't understand how the new coordinates became $(x+y,y-x)$ .
If I apply the formula for rotation of axes I'll get $sqrt{2}$ $(x-y, x+y) $.










share|cite|improve this question















I was reading a book in which it is mentioned that:



Rotate coordinate axes by $45$ degrees so that a point $(x,y)$ becomes $(x+y,y-x)$ .



Here is image 1



Here is image 2



I can't understand how the new coordinates became $(x+y,y-x)$ .
If I apply the formula for rotation of axes I'll get $sqrt{2}$ $(x-y, x+y) $.







geometry rotations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 9:53









dmtri

1,4301521




1,4301521










asked Dec 31 '18 at 6:08









Ayaz S Imran

285




285












  • I agree with you. The distance $BC$ before the rotation is $sqrt {13}$ while it is $sqrt {26}$ afterwards. I think they stretch it back afterwards .
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:21










  • Yes, your are right. But, please explain the context.
    – Martín Vacas Vignolo
    Dec 31 '18 at 6:21










  • @MohammadZuhairKhan I've uploaded the whole section. Can you look something now?
    – Ayaz S Imran
    Dec 31 '18 at 6:25






  • 3




    @MartínVacasVignolo the new coordinates after rotation of the cordinates at an angle f for a point (x,y){x = Xcos(f) - Ysin(f)}{y = Xcos(f) + Ysin(f)}
    – Ayaz S Imran
    Dec 31 '18 at 6:27










  • youtube.com/watch?v=BPgq2AudoEo if you want more details.
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:31


















  • I agree with you. The distance $BC$ before the rotation is $sqrt {13}$ while it is $sqrt {26}$ afterwards. I think they stretch it back afterwards .
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:21










  • Yes, your are right. But, please explain the context.
    – Martín Vacas Vignolo
    Dec 31 '18 at 6:21










  • @MohammadZuhairKhan I've uploaded the whole section. Can you look something now?
    – Ayaz S Imran
    Dec 31 '18 at 6:25






  • 3




    @MartínVacasVignolo the new coordinates after rotation of the cordinates at an angle f for a point (x,y){x = Xcos(f) - Ysin(f)}{y = Xcos(f) + Ysin(f)}
    – Ayaz S Imran
    Dec 31 '18 at 6:27










  • youtube.com/watch?v=BPgq2AudoEo if you want more details.
    – Mohammad Zuhair Khan
    Dec 31 '18 at 6:31
















I agree with you. The distance $BC$ before the rotation is $sqrt {13}$ while it is $sqrt {26}$ afterwards. I think they stretch it back afterwards .
– Mohammad Zuhair Khan
Dec 31 '18 at 6:21




I agree with you. The distance $BC$ before the rotation is $sqrt {13}$ while it is $sqrt {26}$ afterwards. I think they stretch it back afterwards .
– Mohammad Zuhair Khan
Dec 31 '18 at 6:21












Yes, your are right. But, please explain the context.
– Martín Vacas Vignolo
Dec 31 '18 at 6:21




Yes, your are right. But, please explain the context.
– Martín Vacas Vignolo
Dec 31 '18 at 6:21












@MohammadZuhairKhan I've uploaded the whole section. Can you look something now?
– Ayaz S Imran
Dec 31 '18 at 6:25




@MohammadZuhairKhan I've uploaded the whole section. Can you look something now?
– Ayaz S Imran
Dec 31 '18 at 6:25




3




3




@MartínVacasVignolo the new coordinates after rotation of the cordinates at an angle f for a point (x,y){x = Xcos(f) - Ysin(f)}{y = Xcos(f) + Ysin(f)}
– Ayaz S Imran
Dec 31 '18 at 6:27




@MartínVacasVignolo the new coordinates after rotation of the cordinates at an angle f for a point (x,y){x = Xcos(f) - Ysin(f)}{y = Xcos(f) + Ysin(f)}
– Ayaz S Imran
Dec 31 '18 at 6:27












youtube.com/watch?v=BPgq2AudoEo if you want more details.
– Mohammad Zuhair Khan
Dec 31 '18 at 6:31




youtube.com/watch?v=BPgq2AudoEo if you want more details.
– Mohammad Zuhair Khan
Dec 31 '18 at 6:31










4 Answers
4






active

oldest

votes


















5














As noted, that transformation isn't an isometry (in the Euclidean metric); it's a $45^circ$ rotation composed with a dilation by $sqrt{2}$.



But then, looking at the pictures, we're not looking at Euclidean geometry - we're instead looking at the distances derived from the 1-norm $|(x,y)|_1=|x|+|y|$ and the $infty$-norm $|(x,y)|=max(|x|,|y|)$. From that perspective, we're just fine leaving that scale factor in so the transformation has rational coefficients. The formula for $T^{-1}$ will have factors of $frac12$, and that's just not a big deal.



Let $T(x,y)=(x+y,y-x)$. The linked pictures say that $|T(v)|_{infty}=|v|_1$. We also get $|T(v)|_1=2|v|_{infty}$. Taking two steps, $T^2(x,y)=(2y,-2x)$, which doubles both norms. What this is saying is that the geometry we get from the $1$-norm and the geometry we get from the $infty$-norm are isomorphic - because this transformation $T$ transforms one distance into the other. Using the irrational version that's an isometry in the Euclidean metric would only make things worse by introducing a factor of $sqrt{2}$ to the distances we're looking at. Pass.






share|cite|improve this answer























  • The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
    – John Douma
    Dec 31 '18 at 15:58










  • No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
    – jmerry
    Dec 31 '18 at 21:13










  • We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
    – John Douma
    Dec 31 '18 at 21:19










  • No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
    – jmerry
    Dec 31 '18 at 22:34










  • That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
    – John Douma
    Dec 31 '18 at 23:17





















1














You are rotating the coordinate axes, not the points. Let $(x,y)$ be a point in the plane. Then $(x,y)=x(1,0)+y(0,1)$. A rotation of the coordinates by $45^{circ}$ takes $(1,0)$ to $(frac{1}{sqrt{2}},frac{1}{sqrt{2}})$ and $(0,1)$ to $(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})$.



To find the new coordinates $(x',y')$ we must solve $$x'(frac{1}{sqrt{2}},frac{1}{sqrt{2}})+y'(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})=(x,y)$$



Grouping terms on the left of the equation and equating coordinates gives $$x'-y'=sqrt{2}x$$ $$x'+y'=sqrt{2}y$$



The $sqrt{2}$ corresponds to the length of the diagonal of a unit square which is $2$ in taxicab geometry. Replacing $sqrt{2}$ with $2$ in out system of equations gives us $$x'-y'=2x$$ $$x'+y'=2y$$



We can solve this system for $x'$ and $y'$ to obtain $$x'=x+y$$ $$y'=y-x$$.



Edit:



I see many saying that this transformation is a rotation followed by a dilation but this is not accurate. A rotation must preserve length. In this case the length is the taxicab length. Take $(1,0)$. It has a taxicab length of $1$ from the origin. If we rotate this by $45^{circ}$ under the standard metric we get $(frac{1}{sqrt{2}}, frac{1}{sqrt{2}})$. But the same rotation under the taxicab metric must give $(frac{1}{2},frac{1}{2})$ to preserve the length of the rotated vector.



My proof above can be confusing because I don't make that adjustment until after the rotation with respect to the standard metric.






share|cite|improve this answer























  • The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
    – jmerry
    Jan 1 at 2:23










  • @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
    – John Douma
    Jan 1 at 3:30










  • No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
    – jmerry
    Jan 1 at 4:00










  • @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
    – John Douma
    Jan 1 at 15:33



















0














I'd think that the author was more concerned to the underlayed square lattice grid, as being displayed in your linked pictures. That is, integer coordinates ought be transformed into integer coordinates only, and not into something irrational, involving a sqrt(2) factor.



--- rk






share|cite|improve this answer





























    0














    In this text, "rotate" is used in a loose sense, as it is actually combined with a scaling by $sqrt2$, to keep integer coordinates.





    To rotate is the verb that corresponds to rotation. There is no verb for "similarity transformation".






    share|cite|improve this answer





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057471%2frotation-of-axes-by-45-degrees%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      As noted, that transformation isn't an isometry (in the Euclidean metric); it's a $45^circ$ rotation composed with a dilation by $sqrt{2}$.



      But then, looking at the pictures, we're not looking at Euclidean geometry - we're instead looking at the distances derived from the 1-norm $|(x,y)|_1=|x|+|y|$ and the $infty$-norm $|(x,y)|=max(|x|,|y|)$. From that perspective, we're just fine leaving that scale factor in so the transformation has rational coefficients. The formula for $T^{-1}$ will have factors of $frac12$, and that's just not a big deal.



      Let $T(x,y)=(x+y,y-x)$. The linked pictures say that $|T(v)|_{infty}=|v|_1$. We also get $|T(v)|_1=2|v|_{infty}$. Taking two steps, $T^2(x,y)=(2y,-2x)$, which doubles both norms. What this is saying is that the geometry we get from the $1$-norm and the geometry we get from the $infty$-norm are isomorphic - because this transformation $T$ transforms one distance into the other. Using the irrational version that's an isometry in the Euclidean metric would only make things worse by introducing a factor of $sqrt{2}$ to the distances we're looking at. Pass.






      share|cite|improve this answer























      • The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
        – John Douma
        Dec 31 '18 at 15:58










      • No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
        – jmerry
        Dec 31 '18 at 21:13










      • We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
        – John Douma
        Dec 31 '18 at 21:19










      • No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
        – jmerry
        Dec 31 '18 at 22:34










      • That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
        – John Douma
        Dec 31 '18 at 23:17


















      5














      As noted, that transformation isn't an isometry (in the Euclidean metric); it's a $45^circ$ rotation composed with a dilation by $sqrt{2}$.



      But then, looking at the pictures, we're not looking at Euclidean geometry - we're instead looking at the distances derived from the 1-norm $|(x,y)|_1=|x|+|y|$ and the $infty$-norm $|(x,y)|=max(|x|,|y|)$. From that perspective, we're just fine leaving that scale factor in so the transformation has rational coefficients. The formula for $T^{-1}$ will have factors of $frac12$, and that's just not a big deal.



      Let $T(x,y)=(x+y,y-x)$. The linked pictures say that $|T(v)|_{infty}=|v|_1$. We also get $|T(v)|_1=2|v|_{infty}$. Taking two steps, $T^2(x,y)=(2y,-2x)$, which doubles both norms. What this is saying is that the geometry we get from the $1$-norm and the geometry we get from the $infty$-norm are isomorphic - because this transformation $T$ transforms one distance into the other. Using the irrational version that's an isometry in the Euclidean metric would only make things worse by introducing a factor of $sqrt{2}$ to the distances we're looking at. Pass.






      share|cite|improve this answer























      • The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
        – John Douma
        Dec 31 '18 at 15:58










      • No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
        – jmerry
        Dec 31 '18 at 21:13










      • We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
        – John Douma
        Dec 31 '18 at 21:19










      • No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
        – jmerry
        Dec 31 '18 at 22:34










      • That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
        – John Douma
        Dec 31 '18 at 23:17
















      5












      5








      5






      As noted, that transformation isn't an isometry (in the Euclidean metric); it's a $45^circ$ rotation composed with a dilation by $sqrt{2}$.



      But then, looking at the pictures, we're not looking at Euclidean geometry - we're instead looking at the distances derived from the 1-norm $|(x,y)|_1=|x|+|y|$ and the $infty$-norm $|(x,y)|=max(|x|,|y|)$. From that perspective, we're just fine leaving that scale factor in so the transformation has rational coefficients. The formula for $T^{-1}$ will have factors of $frac12$, and that's just not a big deal.



      Let $T(x,y)=(x+y,y-x)$. The linked pictures say that $|T(v)|_{infty}=|v|_1$. We also get $|T(v)|_1=2|v|_{infty}$. Taking two steps, $T^2(x,y)=(2y,-2x)$, which doubles both norms. What this is saying is that the geometry we get from the $1$-norm and the geometry we get from the $infty$-norm are isomorphic - because this transformation $T$ transforms one distance into the other. Using the irrational version that's an isometry in the Euclidean metric would only make things worse by introducing a factor of $sqrt{2}$ to the distances we're looking at. Pass.






      share|cite|improve this answer














      As noted, that transformation isn't an isometry (in the Euclidean metric); it's a $45^circ$ rotation composed with a dilation by $sqrt{2}$.



      But then, looking at the pictures, we're not looking at Euclidean geometry - we're instead looking at the distances derived from the 1-norm $|(x,y)|_1=|x|+|y|$ and the $infty$-norm $|(x,y)|=max(|x|,|y|)$. From that perspective, we're just fine leaving that scale factor in so the transformation has rational coefficients. The formula for $T^{-1}$ will have factors of $frac12$, and that's just not a big deal.



      Let $T(x,y)=(x+y,y-x)$. The linked pictures say that $|T(v)|_{infty}=|v|_1$. We also get $|T(v)|_1=2|v|_{infty}$. Taking two steps, $T^2(x,y)=(2y,-2x)$, which doubles both norms. What this is saying is that the geometry we get from the $1$-norm and the geometry we get from the $infty$-norm are isomorphic - because this transformation $T$ transforms one distance into the other. Using the irrational version that's an isometry in the Euclidean metric would only make things worse by introducing a factor of $sqrt{2}$ to the distances we're looking at. Pass.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 31 '18 at 21:15

























      answered Dec 31 '18 at 8:09









      jmerry

      2,416211




      2,416211












      • The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
        – John Douma
        Dec 31 '18 at 15:58










      • No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
        – jmerry
        Dec 31 '18 at 21:13










      • We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
        – John Douma
        Dec 31 '18 at 21:19










      • No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
        – jmerry
        Dec 31 '18 at 22:34










      • That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
        – John Douma
        Dec 31 '18 at 23:17




















      • The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
        – John Douma
        Dec 31 '18 at 15:58










      • No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
        – jmerry
        Dec 31 '18 at 21:13










      • We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
        – John Douma
        Dec 31 '18 at 21:19










      • No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
        – jmerry
        Dec 31 '18 at 22:34










      • That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
        – John Douma
        Dec 31 '18 at 23:17


















      The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
      – John Douma
      Dec 31 '18 at 15:58




      The transformation is an isometry. The length that is preserved is the taxicab length. For example, $(1,0)$ gets mapped to $(frac{1}{2}, frac{1}{2})$ under rotation which preserves the taxicab length of the vector.
      – John Douma
      Dec 31 '18 at 15:58












      No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
      – jmerry
      Dec 31 '18 at 21:13




      No, @John Douma, it's not an isometry. The taxicab distance is not preserved - it's sent to that other $infty$-distance.
      – jmerry
      Dec 31 '18 at 21:13












      We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
      – John Douma
      Dec 31 '18 at 21:19




      We are not rotating our grid. We are rotating our coordinates. Draw a grid and place the origin at the southwest corner of a building. There are only two ways to get to the northeast corner no matter how we describe it. No distances have changed. If you rotate the x axis by $45^{circ}$ you describe the northeast corner as $(2,0)$ instead of $(1,1)$ but all distances are unchanged.
      – John Douma
      Dec 31 '18 at 21:19












      No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
      – jmerry
      Dec 31 '18 at 22:34




      No, not all distances are preserved. In your example, what about the distance to an adjacent corner of the building? Recall also that applying $T$ twice doubles all distances.
      – jmerry
      Dec 31 '18 at 22:34












      That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
      – John Douma
      Dec 31 '18 at 23:17






      That distance is $1$. The geometry has not changed. You cannot cut through a building on its diagonal just because you changed its description. With the new coordinate system, $(1,0)$ is described by $(1,-1)$ and is a distance of $1$ away. We are still restricted to travel along the grid lines even if they are diagonal in our new coordinate system.
      – John Douma
      Dec 31 '18 at 23:17













      1














      You are rotating the coordinate axes, not the points. Let $(x,y)$ be a point in the plane. Then $(x,y)=x(1,0)+y(0,1)$. A rotation of the coordinates by $45^{circ}$ takes $(1,0)$ to $(frac{1}{sqrt{2}},frac{1}{sqrt{2}})$ and $(0,1)$ to $(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})$.



      To find the new coordinates $(x',y')$ we must solve $$x'(frac{1}{sqrt{2}},frac{1}{sqrt{2}})+y'(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})=(x,y)$$



      Grouping terms on the left of the equation and equating coordinates gives $$x'-y'=sqrt{2}x$$ $$x'+y'=sqrt{2}y$$



      The $sqrt{2}$ corresponds to the length of the diagonal of a unit square which is $2$ in taxicab geometry. Replacing $sqrt{2}$ with $2$ in out system of equations gives us $$x'-y'=2x$$ $$x'+y'=2y$$



      We can solve this system for $x'$ and $y'$ to obtain $$x'=x+y$$ $$y'=y-x$$.



      Edit:



      I see many saying that this transformation is a rotation followed by a dilation but this is not accurate. A rotation must preserve length. In this case the length is the taxicab length. Take $(1,0)$. It has a taxicab length of $1$ from the origin. If we rotate this by $45^{circ}$ under the standard metric we get $(frac{1}{sqrt{2}}, frac{1}{sqrt{2}})$. But the same rotation under the taxicab metric must give $(frac{1}{2},frac{1}{2})$ to preserve the length of the rotated vector.



      My proof above can be confusing because I don't make that adjustment until after the rotation with respect to the standard metric.






      share|cite|improve this answer























      • The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
        – jmerry
        Jan 1 at 2:23










      • @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
        – John Douma
        Jan 1 at 3:30










      • No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
        – jmerry
        Jan 1 at 4:00










      • @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
        – John Douma
        Jan 1 at 15:33
















      1














      You are rotating the coordinate axes, not the points. Let $(x,y)$ be a point in the plane. Then $(x,y)=x(1,0)+y(0,1)$. A rotation of the coordinates by $45^{circ}$ takes $(1,0)$ to $(frac{1}{sqrt{2}},frac{1}{sqrt{2}})$ and $(0,1)$ to $(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})$.



      To find the new coordinates $(x',y')$ we must solve $$x'(frac{1}{sqrt{2}},frac{1}{sqrt{2}})+y'(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})=(x,y)$$



      Grouping terms on the left of the equation and equating coordinates gives $$x'-y'=sqrt{2}x$$ $$x'+y'=sqrt{2}y$$



      The $sqrt{2}$ corresponds to the length of the diagonal of a unit square which is $2$ in taxicab geometry. Replacing $sqrt{2}$ with $2$ in out system of equations gives us $$x'-y'=2x$$ $$x'+y'=2y$$



      We can solve this system for $x'$ and $y'$ to obtain $$x'=x+y$$ $$y'=y-x$$.



      Edit:



      I see many saying that this transformation is a rotation followed by a dilation but this is not accurate. A rotation must preserve length. In this case the length is the taxicab length. Take $(1,0)$. It has a taxicab length of $1$ from the origin. If we rotate this by $45^{circ}$ under the standard metric we get $(frac{1}{sqrt{2}}, frac{1}{sqrt{2}})$. But the same rotation under the taxicab metric must give $(frac{1}{2},frac{1}{2})$ to preserve the length of the rotated vector.



      My proof above can be confusing because I don't make that adjustment until after the rotation with respect to the standard metric.






      share|cite|improve this answer























      • The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
        – jmerry
        Jan 1 at 2:23










      • @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
        – John Douma
        Jan 1 at 3:30










      • No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
        – jmerry
        Jan 1 at 4:00










      • @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
        – John Douma
        Jan 1 at 15:33














      1












      1








      1






      You are rotating the coordinate axes, not the points. Let $(x,y)$ be a point in the plane. Then $(x,y)=x(1,0)+y(0,1)$. A rotation of the coordinates by $45^{circ}$ takes $(1,0)$ to $(frac{1}{sqrt{2}},frac{1}{sqrt{2}})$ and $(0,1)$ to $(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})$.



      To find the new coordinates $(x',y')$ we must solve $$x'(frac{1}{sqrt{2}},frac{1}{sqrt{2}})+y'(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})=(x,y)$$



      Grouping terms on the left of the equation and equating coordinates gives $$x'-y'=sqrt{2}x$$ $$x'+y'=sqrt{2}y$$



      The $sqrt{2}$ corresponds to the length of the diagonal of a unit square which is $2$ in taxicab geometry. Replacing $sqrt{2}$ with $2$ in out system of equations gives us $$x'-y'=2x$$ $$x'+y'=2y$$



      We can solve this system for $x'$ and $y'$ to obtain $$x'=x+y$$ $$y'=y-x$$.



      Edit:



      I see many saying that this transformation is a rotation followed by a dilation but this is not accurate. A rotation must preserve length. In this case the length is the taxicab length. Take $(1,0)$. It has a taxicab length of $1$ from the origin. If we rotate this by $45^{circ}$ under the standard metric we get $(frac{1}{sqrt{2}}, frac{1}{sqrt{2}})$. But the same rotation under the taxicab metric must give $(frac{1}{2},frac{1}{2})$ to preserve the length of the rotated vector.



      My proof above can be confusing because I don't make that adjustment until after the rotation with respect to the standard metric.






      share|cite|improve this answer














      You are rotating the coordinate axes, not the points. Let $(x,y)$ be a point in the plane. Then $(x,y)=x(1,0)+y(0,1)$. A rotation of the coordinates by $45^{circ}$ takes $(1,0)$ to $(frac{1}{sqrt{2}},frac{1}{sqrt{2}})$ and $(0,1)$ to $(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})$.



      To find the new coordinates $(x',y')$ we must solve $$x'(frac{1}{sqrt{2}},frac{1}{sqrt{2}})+y'(frac{-1}{sqrt{2}},frac{1}{sqrt{2}})=(x,y)$$



      Grouping terms on the left of the equation and equating coordinates gives $$x'-y'=sqrt{2}x$$ $$x'+y'=sqrt{2}y$$



      The $sqrt{2}$ corresponds to the length of the diagonal of a unit square which is $2$ in taxicab geometry. Replacing $sqrt{2}$ with $2$ in out system of equations gives us $$x'-y'=2x$$ $$x'+y'=2y$$



      We can solve this system for $x'$ and $y'$ to obtain $$x'=x+y$$ $$y'=y-x$$.



      Edit:



      I see many saying that this transformation is a rotation followed by a dilation but this is not accurate. A rotation must preserve length. In this case the length is the taxicab length. Take $(1,0)$. It has a taxicab length of $1$ from the origin. If we rotate this by $45^{circ}$ under the standard metric we get $(frac{1}{sqrt{2}}, frac{1}{sqrt{2}})$. But the same rotation under the taxicab metric must give $(frac{1}{2},frac{1}{2})$ to preserve the length of the rotated vector.



      My proof above can be confusing because I don't make that adjustment until after the rotation with respect to the standard metric.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 31 '18 at 16:06

























      answered Dec 31 '18 at 9:03









      John Douma

      5,40211319




      5,40211319












      • The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
        – jmerry
        Jan 1 at 2:23










      • @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
        – John Douma
        Jan 1 at 3:30










      • No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
        – jmerry
        Jan 1 at 4:00










      • @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
        – John Douma
        Jan 1 at 15:33


















      • The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
        – jmerry
        Jan 1 at 2:23










      • @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
        – John Douma
        Jan 1 at 3:30










      • No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
        – jmerry
        Jan 1 at 4:00










      • @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
        – John Douma
        Jan 1 at 15:33
















      The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
      – jmerry
      Jan 1 at 2:23




      The paragraph edited in is wrong; this transformation is not an isometry in the taxicab metric, or indeed in any one metric. Every isometry of taxicab geometry must send vertical and horizontal segments to vertical and horizontal segments, as they can be characterized as the only segments with unique midpoints.
      – jmerry
      Jan 1 at 2:23












      @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
      – John Douma
      Jan 1 at 3:30




      @jmerry A coordinate rotation changes how we describe points. It doesn't change the geometry of the grid. You do agree that in our new coordinate systems the grid lines are diagonal, right?
      – John Douma
      Jan 1 at 3:30












      No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
      – jmerry
      Jan 1 at 4:00




      No, I don't agree. The material linked in the question doesn't agree with you either - no rotated grid lines there. It's a new distance function, and thus a new geometry, which just happens to be related to the old one.
      – jmerry
      Jan 1 at 4:00












      @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
      – John Douma
      Jan 1 at 15:33




      @jmerry Please read the last part of the second picture carefully. We compute distances differently in the rotated coordinate system. The example computes the difference between two points. The author explicitly says we have two ways to compute the $same$ distance. Look at the first picture where the distance between $B$ and $C$ is shown. Now look at the rotated picture and tell me what the distance between the two is. If you think it is $6$, go back to the second picture and read it again.
      – John Douma
      Jan 1 at 15:33











      0














      I'd think that the author was more concerned to the underlayed square lattice grid, as being displayed in your linked pictures. That is, integer coordinates ought be transformed into integer coordinates only, and not into something irrational, involving a sqrt(2) factor.



      --- rk






      share|cite|improve this answer


























        0














        I'd think that the author was more concerned to the underlayed square lattice grid, as being displayed in your linked pictures. That is, integer coordinates ought be transformed into integer coordinates only, and not into something irrational, involving a sqrt(2) factor.



        --- rk






        share|cite|improve this answer
























          0












          0








          0






          I'd think that the author was more concerned to the underlayed square lattice grid, as being displayed in your linked pictures. That is, integer coordinates ought be transformed into integer coordinates only, and not into something irrational, involving a sqrt(2) factor.



          --- rk






          share|cite|improve this answer












          I'd think that the author was more concerned to the underlayed square lattice grid, as being displayed in your linked pictures. That is, integer coordinates ought be transformed into integer coordinates only, and not into something irrational, involving a sqrt(2) factor.



          --- rk







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 10:44









          Dr. Richard Klitzing

          1,38116




          1,38116























              0














              In this text, "rotate" is used in a loose sense, as it is actually combined with a scaling by $sqrt2$, to keep integer coordinates.





              To rotate is the verb that corresponds to rotation. There is no verb for "similarity transformation".






              share|cite|improve this answer


























                0














                In this text, "rotate" is used in a loose sense, as it is actually combined with a scaling by $sqrt2$, to keep integer coordinates.





                To rotate is the verb that corresponds to rotation. There is no verb for "similarity transformation".






                share|cite|improve this answer
























                  0












                  0








                  0






                  In this text, "rotate" is used in a loose sense, as it is actually combined with a scaling by $sqrt2$, to keep integer coordinates.





                  To rotate is the verb that corresponds to rotation. There is no verb for "similarity transformation".






                  share|cite|improve this answer












                  In this text, "rotate" is used in a loose sense, as it is actually combined with a scaling by $sqrt2$, to keep integer coordinates.





                  To rotate is the verb that corresponds to rotation. There is no verb for "similarity transformation".







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 11:19









                  Yves Daoust

                  124k671222




                  124k671222






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057471%2frotation-of-axes-by-45-degrees%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Human spaceflight

                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                      張江高科駅