Orthogonal Matrix with a specific row












0














I have an assignment with the following question:



Does an Orthogonal Matrix exist such that its first row consists of the



 following values:


($1$/$sqrt{3}$, $1$/$sqrt{3}$, $1$/$sqrt{3}$)



If there is, find one.


I know I can solve this question with the Gram Schmidt algorithm, but it includes a lot of complicated calculations.



Is there any other way to prove this statement without the Gram Schmidt algorithm?



Thanks,



Alan






linear-algebra orthonormal orthogonality







share|cite|improve this question
























  • There is another way. You should be able to show that each row (column) of an orthogonal matrix is orthonormal to the other rows (columns) of the matrix. That fact should be enough to answer the original question.
    – LouisB
    Nov 23 '14 at 6:32










  • @Louis, can you please elaborate, as I amnot sure how to prove that? Thank you in advance.
    – Alan
    Nov 23 '14 at 6:34












  • I misread the question! I managed to read $1, {sqrt{3}}$ instead of $1/{sqrt{3}}$. Sorry about that.
    – LouisB
    Nov 23 '14 at 7:01










  • Similar: math.stackexchange.com/questions/2241414/….
    – StubbornAtom
    May 7 '17 at 10:56










  • An orthogonal matrix will have mutually perpendicular rows of unit vectors so as long as the vector is a unit vector you're have sufficient conditions for the existence of such a matrix.
    – CyclotomicField
    Jul 2 '18 at 4:31
















0














I have an assignment with the following question:



Does an Orthogonal Matrix exist such that its first row consists of the



 following values:


($1$/$sqrt{3}$, $1$/$sqrt{3}$, $1$/$sqrt{3}$)



If there is, find one.


I know I can solve this question with the Gram Schmidt algorithm, but it includes a lot of complicated calculations.



Is there any other way to prove this statement without the Gram Schmidt algorithm?



Thanks,



Alan






linear-algebra orthonormal orthogonality







share|cite|improve this question
























  • There is another way. You should be able to show that each row (column) of an orthogonal matrix is orthonormal to the other rows (columns) of the matrix. That fact should be enough to answer the original question.
    – LouisB
    Nov 23 '14 at 6:32










  • @Louis, can you please elaborate, as I amnot sure how to prove that? Thank you in advance.
    – Alan
    Nov 23 '14 at 6:34












  • I misread the question! I managed to read $1, {sqrt{3}}$ instead of $1/{sqrt{3}}$. Sorry about that.
    – LouisB
    Nov 23 '14 at 7:01










  • Similar: math.stackexchange.com/questions/2241414/….
    – StubbornAtom
    May 7 '17 at 10:56










  • An orthogonal matrix will have mutually perpendicular rows of unit vectors so as long as the vector is a unit vector you're have sufficient conditions for the existence of such a matrix.
    – CyclotomicField
    Jul 2 '18 at 4:31














0












0








0


1





I have an assignment with the following question:



Does an Orthogonal Matrix exist such that its first row consists of the



 following values:


($1$/$sqrt{3}$, $1$/$sqrt{3}$, $1$/$sqrt{3}$)



If there is, find one.


I know I can solve this question with the Gram Schmidt algorithm, but it includes a lot of complicated calculations.



Is there any other way to prove this statement without the Gram Schmidt algorithm?



Thanks,



Alan






linear-algebra orthonormal orthogonality







share|cite|improve this question















I have an assignment with the following question:



Does an Orthogonal Matrix exist such that its first row consists of the



 following values:


($1$/$sqrt{3}$, $1$/$sqrt{3}$, $1$/$sqrt{3}$)



If there is, find one.


I know I can solve this question with the Gram Schmidt algorithm, but it includes a lot of complicated calculations.



Is there any other way to prove this statement without the Gram Schmidt algorithm?



Thanks,



Alan






linear-algebra orthonormal orthogonality




linear-algebra orthonormal orthogonality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '14 at 6:27

























asked Nov 23 '14 at 6:13









Alan

1,3671021




1,3671021












  • There is another way. You should be able to show that each row (column) of an orthogonal matrix is orthonormal to the other rows (columns) of the matrix. That fact should be enough to answer the original question.
    – LouisB
    Nov 23 '14 at 6:32










  • @Louis, can you please elaborate, as I amnot sure how to prove that? Thank you in advance.
    – Alan
    Nov 23 '14 at 6:34












  • I misread the question! I managed to read $1, {sqrt{3}}$ instead of $1/{sqrt{3}}$. Sorry about that.
    – LouisB
    Nov 23 '14 at 7:01










  • Similar: math.stackexchange.com/questions/2241414/….
    – StubbornAtom
    May 7 '17 at 10:56










  • An orthogonal matrix will have mutually perpendicular rows of unit vectors so as long as the vector is a unit vector you're have sufficient conditions for the existence of such a matrix.
    – CyclotomicField
    Jul 2 '18 at 4:31


















  • There is another way. You should be able to show that each row (column) of an orthogonal matrix is orthonormal to the other rows (columns) of the matrix. That fact should be enough to answer the original question.
    – LouisB
    Nov 23 '14 at 6:32










  • @Louis, can you please elaborate, as I amnot sure how to prove that? Thank you in advance.
    – Alan
    Nov 23 '14 at 6:34












  • I misread the question! I managed to read $1, {sqrt{3}}$ instead of $1/{sqrt{3}}$. Sorry about that.
    – LouisB
    Nov 23 '14 at 7:01










  • Similar: math.stackexchange.com/questions/2241414/….
    – StubbornAtom
    May 7 '17 at 10:56










  • An orthogonal matrix will have mutually perpendicular rows of unit vectors so as long as the vector is a unit vector you're have sufficient conditions for the existence of such a matrix.
    – CyclotomicField
    Jul 2 '18 at 4:31
















There is another way. You should be able to show that each row (column) of an orthogonal matrix is orthonormal to the other rows (columns) of the matrix. That fact should be enough to answer the original question.
– LouisB
Nov 23 '14 at 6:32




There is another way. You should be able to show that each row (column) of an orthogonal matrix is orthonormal to the other rows (columns) of the matrix. That fact should be enough to answer the original question.
– LouisB
Nov 23 '14 at 6:32












@Louis, can you please elaborate, as I amnot sure how to prove that? Thank you in advance.
– Alan
Nov 23 '14 at 6:34






@Louis, can you please elaborate, as I amnot sure how to prove that? Thank you in advance.
– Alan
Nov 23 '14 at 6:34














I misread the question! I managed to read $1, {sqrt{3}}$ instead of $1/{sqrt{3}}$. Sorry about that.
– LouisB
Nov 23 '14 at 7:01




I misread the question! I managed to read $1, {sqrt{3}}$ instead of $1/{sqrt{3}}$. Sorry about that.
– LouisB
Nov 23 '14 at 7:01












Similar: math.stackexchange.com/questions/2241414/….
– StubbornAtom
May 7 '17 at 10:56




Similar: math.stackexchange.com/questions/2241414/….
– StubbornAtom
May 7 '17 at 10:56












An orthogonal matrix will have mutually perpendicular rows of unit vectors so as long as the vector is a unit vector you're have sufficient conditions for the existence of such a matrix.
– CyclotomicField
Jul 2 '18 at 4:31




An orthogonal matrix will have mutually perpendicular rows of unit vectors so as long as the vector is a unit vector you're have sufficient conditions for the existence of such a matrix.
– CyclotomicField
Jul 2 '18 at 4:31










3 Answers
3






active

oldest

votes


















0














You need to choose two vectors which are orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ and make sure they are also orthogonal to each other.



Vectors orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ lie in the plane $x+y+z=0$.



For example choose $(1,-1,0)$ to be the second row of your matrix.



Now you need a third vector which satisfies the equations:



$$x+y+z=0 ~~~ text{[is orthogonal to}~left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)]$$
$$x-y=0 ~~~ text{[is orthogonal to}~(1,-1,0)]$$



Any such vector can form the third row of your orthogonal matrix.






share|cite|improve this answer





















  • One way to find a 3rd vector is to take the cross product of the first two vectors.
    – Gerry Myerson
    Nov 23 '14 at 10:57



















0














i answered a similar question before, but i don't know how to make a link to that.



so i will repeat the way to find a symmetric reflection matrix whose row is your unit vector. reflect $((1,1,1)^T/sqrt 3$ to $(1,0,0)^T$ on the mirror in the direction of their bisector $a = (1 + sqrt 3, 1, 1)^T.$ form the matrix $q = 2aa^T/a^Ta - I;$ that will satisfy your requirement.






share|cite|improve this answer





























    0














    $(frac1{sqrt3},frac1{sqrt3},frac1{sqrt3})$ is a unit vector, so you're off to a good start. (That is, there is such a matrix.)



    Next just choose a perpendicular vector, and normalize it. So, say, $(frac1{sqrt2},-frac1{sqrt2},0)$.



    Finally you need a vector perpendicular to the first two, and again of unit length. The cross-product will produce such a vector. Or, looking at the first two, I think I can produce one out of the blue. Namely, $(frac1{sqrt6},frac1{sqrt6},-frac2{sqrt6})$. You can easily check that the dot products with the other two vectors are zero. And it has unit length.



    So one such matrix is: $$begin{pmatrix}frac1{sqrt3}& frac1{sqrt3}& frac1{sqrt3}\frac1{sqrt2}&-frac1{sqrt2}&0\frac1{sqrt6}&frac1{sqrt6}&-frac2{sqrt6}end{pmatrix}$$.






    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%2f1034698%2forthogonal-matrix-with-a-specific-row%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      You need to choose two vectors which are orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ and make sure they are also orthogonal to each other.



      Vectors orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ lie in the plane $x+y+z=0$.



      For example choose $(1,-1,0)$ to be the second row of your matrix.



      Now you need a third vector which satisfies the equations:



      $$x+y+z=0 ~~~ text{[is orthogonal to}~left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)]$$
      $$x-y=0 ~~~ text{[is orthogonal to}~(1,-1,0)]$$



      Any such vector can form the third row of your orthogonal matrix.






      share|cite|improve this answer





















      • One way to find a 3rd vector is to take the cross product of the first two vectors.
        – Gerry Myerson
        Nov 23 '14 at 10:57
















      0














      You need to choose two vectors which are orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ and make sure they are also orthogonal to each other.



      Vectors orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ lie in the plane $x+y+z=0$.



      For example choose $(1,-1,0)$ to be the second row of your matrix.



      Now you need a third vector which satisfies the equations:



      $$x+y+z=0 ~~~ text{[is orthogonal to}~left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)]$$
      $$x-y=0 ~~~ text{[is orthogonal to}~(1,-1,0)]$$



      Any such vector can form the third row of your orthogonal matrix.






      share|cite|improve this answer





















      • One way to find a 3rd vector is to take the cross product of the first two vectors.
        – Gerry Myerson
        Nov 23 '14 at 10:57














      0












      0








      0






      You need to choose two vectors which are orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ and make sure they are also orthogonal to each other.



      Vectors orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ lie in the plane $x+y+z=0$.



      For example choose $(1,-1,0)$ to be the second row of your matrix.



      Now you need a third vector which satisfies the equations:



      $$x+y+z=0 ~~~ text{[is orthogonal to}~left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)]$$
      $$x-y=0 ~~~ text{[is orthogonal to}~(1,-1,0)]$$



      Any such vector can form the third row of your orthogonal matrix.






      share|cite|improve this answer












      You need to choose two vectors which are orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ and make sure they are also orthogonal to each other.



      Vectors orthogonal to $left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)$ lie in the plane $x+y+z=0$.



      For example choose $(1,-1,0)$ to be the second row of your matrix.



      Now you need a third vector which satisfies the equations:



      $$x+y+z=0 ~~~ text{[is orthogonal to}~left(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}}right)]$$
      $$x-y=0 ~~~ text{[is orthogonal to}~(1,-1,0)]$$



      Any such vector can form the third row of your orthogonal matrix.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 23 '14 at 6:39









      RHP

      1,40811525




      1,40811525












      • One way to find a 3rd vector is to take the cross product of the first two vectors.
        – Gerry Myerson
        Nov 23 '14 at 10:57


















      • One way to find a 3rd vector is to take the cross product of the first two vectors.
        – Gerry Myerson
        Nov 23 '14 at 10:57
















      One way to find a 3rd vector is to take the cross product of the first two vectors.
      – Gerry Myerson
      Nov 23 '14 at 10:57




      One way to find a 3rd vector is to take the cross product of the first two vectors.
      – Gerry Myerson
      Nov 23 '14 at 10:57











      0














      i answered a similar question before, but i don't know how to make a link to that.



      so i will repeat the way to find a symmetric reflection matrix whose row is your unit vector. reflect $((1,1,1)^T/sqrt 3$ to $(1,0,0)^T$ on the mirror in the direction of their bisector $a = (1 + sqrt 3, 1, 1)^T.$ form the matrix $q = 2aa^T/a^Ta - I;$ that will satisfy your requirement.






      share|cite|improve this answer


























        0














        i answered a similar question before, but i don't know how to make a link to that.



        so i will repeat the way to find a symmetric reflection matrix whose row is your unit vector. reflect $((1,1,1)^T/sqrt 3$ to $(1,0,0)^T$ on the mirror in the direction of their bisector $a = (1 + sqrt 3, 1, 1)^T.$ form the matrix $q = 2aa^T/a^Ta - I;$ that will satisfy your requirement.






        share|cite|improve this answer
























          0












          0








          0






          i answered a similar question before, but i don't know how to make a link to that.



          so i will repeat the way to find a symmetric reflection matrix whose row is your unit vector. reflect $((1,1,1)^T/sqrt 3$ to $(1,0,0)^T$ on the mirror in the direction of their bisector $a = (1 + sqrt 3, 1, 1)^T.$ form the matrix $q = 2aa^T/a^Ta - I;$ that will satisfy your requirement.






          share|cite|improve this answer












          i answered a similar question before, but i don't know how to make a link to that.



          so i will repeat the way to find a symmetric reflection matrix whose row is your unit vector. reflect $((1,1,1)^T/sqrt 3$ to $(1,0,0)^T$ on the mirror in the direction of their bisector $a = (1 + sqrt 3, 1, 1)^T.$ form the matrix $q = 2aa^T/a^Ta - I;$ that will satisfy your requirement.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '14 at 15:50









          abel

          26.5k11948




          26.5k11948























              0














              $(frac1{sqrt3},frac1{sqrt3},frac1{sqrt3})$ is a unit vector, so you're off to a good start. (That is, there is such a matrix.)



              Next just choose a perpendicular vector, and normalize it. So, say, $(frac1{sqrt2},-frac1{sqrt2},0)$.



              Finally you need a vector perpendicular to the first two, and again of unit length. The cross-product will produce such a vector. Or, looking at the first two, I think I can produce one out of the blue. Namely, $(frac1{sqrt6},frac1{sqrt6},-frac2{sqrt6})$. You can easily check that the dot products with the other two vectors are zero. And it has unit length.



              So one such matrix is: $$begin{pmatrix}frac1{sqrt3}& frac1{sqrt3}& frac1{sqrt3}\frac1{sqrt2}&-frac1{sqrt2}&0\frac1{sqrt6}&frac1{sqrt6}&-frac2{sqrt6}end{pmatrix}$$.






              share|cite|improve this answer


























                0














                $(frac1{sqrt3},frac1{sqrt3},frac1{sqrt3})$ is a unit vector, so you're off to a good start. (That is, there is such a matrix.)



                Next just choose a perpendicular vector, and normalize it. So, say, $(frac1{sqrt2},-frac1{sqrt2},0)$.



                Finally you need a vector perpendicular to the first two, and again of unit length. The cross-product will produce such a vector. Or, looking at the first two, I think I can produce one out of the blue. Namely, $(frac1{sqrt6},frac1{sqrt6},-frac2{sqrt6})$. You can easily check that the dot products with the other two vectors are zero. And it has unit length.



                So one such matrix is: $$begin{pmatrix}frac1{sqrt3}& frac1{sqrt3}& frac1{sqrt3}\frac1{sqrt2}&-frac1{sqrt2}&0\frac1{sqrt6}&frac1{sqrt6}&-frac2{sqrt6}end{pmatrix}$$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  $(frac1{sqrt3},frac1{sqrt3},frac1{sqrt3})$ is a unit vector, so you're off to a good start. (That is, there is such a matrix.)



                  Next just choose a perpendicular vector, and normalize it. So, say, $(frac1{sqrt2},-frac1{sqrt2},0)$.



                  Finally you need a vector perpendicular to the first two, and again of unit length. The cross-product will produce such a vector. Or, looking at the first two, I think I can produce one out of the blue. Namely, $(frac1{sqrt6},frac1{sqrt6},-frac2{sqrt6})$. You can easily check that the dot products with the other two vectors are zero. And it has unit length.



                  So one such matrix is: $$begin{pmatrix}frac1{sqrt3}& frac1{sqrt3}& frac1{sqrt3}\frac1{sqrt2}&-frac1{sqrt2}&0\frac1{sqrt6}&frac1{sqrt6}&-frac2{sqrt6}end{pmatrix}$$.






                  share|cite|improve this answer












                  $(frac1{sqrt3},frac1{sqrt3},frac1{sqrt3})$ is a unit vector, so you're off to a good start. (That is, there is such a matrix.)



                  Next just choose a perpendicular vector, and normalize it. So, say, $(frac1{sqrt2},-frac1{sqrt2},0)$.



                  Finally you need a vector perpendicular to the first two, and again of unit length. The cross-product will produce such a vector. Or, looking at the first two, I think I can produce one out of the blue. Namely, $(frac1{sqrt6},frac1{sqrt6},-frac2{sqrt6})$. You can easily check that the dot products with the other two vectors are zero. And it has unit length.



                  So one such matrix is: $$begin{pmatrix}frac1{sqrt3}& frac1{sqrt3}& frac1{sqrt3}\frac1{sqrt2}&-frac1{sqrt2}&0\frac1{sqrt6}&frac1{sqrt6}&-frac2{sqrt6}end{pmatrix}$$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 7:29









                  Chris Custer

                  10.9k3824




                  10.9k3824






























                      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%2f1034698%2forthogonal-matrix-with-a-specific-row%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

                      Willebadessen

                      Image
                      Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
                      Read more

                      Residenzschloss Arolsen

                      Image
                      Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
                      Read more