Creating a diagonal matrix from from a vector












1












$begingroup$


I try to create a function that create a diagonal matrix in which the diagonal are:



$qquad 1,a,a^2, cdots,a^n,1,a,a^2, cdots,a^n$



Here is my attempt:



nth[n_] := DiagonalMatrix[Table[a^{Mod[k, n]}, {k, 0, 2 n - 1}]];


For some reason and I do not know wh,y the diagonal matrix is just one vector. So it seems DiagonalMatrix does not work as I expect.



Any idea as to where the mistake is in the function I created?






functions matrix table







share|improve this question











$endgroup$

















    1












    $begingroup$


    I try to create a function that create a diagonal matrix in which the diagonal are:



    $qquad 1,a,a^2, cdots,a^n,1,a,a^2, cdots,a^n$



    Here is my attempt:



    nth[n_] := DiagonalMatrix[Table[a^{Mod[k, n]}, {k, 0, 2 n - 1}]];


    For some reason and I do not know wh,y the diagonal matrix is just one vector. So it seems DiagonalMatrix does not work as I expect.



    Any idea as to where the mistake is in the function I created?






    functions matrix table







    share|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I try to create a function that create a diagonal matrix in which the diagonal are:



      $qquad 1,a,a^2, cdots,a^n,1,a,a^2, cdots,a^n$



      Here is my attempt:



      nth[n_] := DiagonalMatrix[Table[a^{Mod[k, n]}, {k, 0, 2 n - 1}]];


      For some reason and I do not know wh,y the diagonal matrix is just one vector. So it seems DiagonalMatrix does not work as I expect.



      Any idea as to where the mistake is in the function I created?






      functions matrix table







      share|improve this question











      $endgroup$




      I try to create a function that create a diagonal matrix in which the diagonal are:



      $qquad 1,a,a^2, cdots,a^n,1,a,a^2, cdots,a^n$



      Here is my attempt:



      nth[n_] := DiagonalMatrix[Table[a^{Mod[k, n]}, {k, 0, 2 n - 1}]];


      For some reason and I do not know wh,y the diagonal matrix is just one vector. So it seems DiagonalMatrix does not work as I expect.



      Any idea as to where the mistake is in the function I created?






      functions matrix table




      functions matrix table






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 27 mins ago









      m_goldberg

      85.9k872196




      85.9k872196










      asked 6 hours ago









      henryhenry

      1346




      1346






















          1 Answer
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          3












          $begingroup$

          Remove the braces around Mod:



          ClearAll[nth]
          
nth[n_] := DiagonalMatrix@Table[a^Mod[k, n], {k, 0, 2 n - 1}]
          

          
nth[5] // MatrixForm // TeXForm



          $left(
          begin{array}{cccccccccc}
          1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & a & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & a^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & a^3 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & a^4 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & a & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & a^2 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^3 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^4 \
          end{array}
          right)$




          Alternatively, you can use SparseArray:



          ClearAll[nth2]
          
nth2[n_] := SparseArray[{k_, k_} :> a^Mod[k, n], {2 n - 1, 2 n - 1}]
          
nth2[5] // MatrixForm // TeXForm



          same result




          or use Band in combination with SparseArray:



          ClearAll[nth3]
          
nth3[n_] := SparseArray[Band[{1, 1}] -> Table[a^Mod[k, n], {k, 0, 2 n - 1}]]
          

          
nth2[5] // MatrixForm // TeXForm



          same result







          share|improve this answer











          $endgroup$













          • $begingroup$
            thanks.. it work perfectly.
            $endgroup$
            – henry
            5 hours ago










          • $begingroup$
            which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
            $endgroup$
            – henry
            5 hours ago








          • 1




            $begingroup$
            @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
            $endgroup$
            – kglr
            5 hours ago










          • $begingroup$
            final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
            $endgroup$
            – henry
            5 hours ago






          • 1




            $begingroup$
            @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
            $endgroup$
            – kglr
            5 hours ago











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          3












          $begingroup$

          Remove the braces around Mod:



          ClearAll[nth]
          
nth[n_] := DiagonalMatrix@Table[a^Mod[k, n], {k, 0, 2 n - 1}]
          

          
nth[5] // MatrixForm // TeXForm



          $left(
          begin{array}{cccccccccc}
          1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & a & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & a^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & a^3 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & a^4 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & a & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & a^2 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^3 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^4 \
          end{array}
          right)$




          Alternatively, you can use SparseArray:



          ClearAll[nth2]
          
nth2[n_] := SparseArray[{k_, k_} :> a^Mod[k, n], {2 n - 1, 2 n - 1}]
          
nth2[5] // MatrixForm // TeXForm



          same result




          or use Band in combination with SparseArray:



          ClearAll[nth3]
          
nth3[n_] := SparseArray[Band[{1, 1}] -> Table[a^Mod[k, n], {k, 0, 2 n - 1}]]
          

          
nth2[5] // MatrixForm // TeXForm



          same result







          share|improve this answer











          $endgroup$













          • $begingroup$
            thanks.. it work perfectly.
            $endgroup$
            – henry
            5 hours ago










          • $begingroup$
            which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
            $endgroup$
            – henry
            5 hours ago








          • 1




            $begingroup$
            @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
            $endgroup$
            – kglr
            5 hours ago










          • $begingroup$
            final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
            $endgroup$
            – henry
            5 hours ago






          • 1




            $begingroup$
            @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
            $endgroup$
            – kglr
            5 hours ago
















          3












          $begingroup$

          Remove the braces around Mod:



          ClearAll[nth]
          
nth[n_] := DiagonalMatrix@Table[a^Mod[k, n], {k, 0, 2 n - 1}]
          

          
nth[5] // MatrixForm // TeXForm



          $left(
          begin{array}{cccccccccc}
          1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & a & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & a^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & a^3 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & a^4 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & a & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & a^2 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^3 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^4 \
          end{array}
          right)$




          Alternatively, you can use SparseArray:



          ClearAll[nth2]
          
nth2[n_] := SparseArray[{k_, k_} :> a^Mod[k, n], {2 n - 1, 2 n - 1}]
          
nth2[5] // MatrixForm // TeXForm



          same result




          or use Band in combination with SparseArray:



          ClearAll[nth3]
          
nth3[n_] := SparseArray[Band[{1, 1}] -> Table[a^Mod[k, n], {k, 0, 2 n - 1}]]
          

          
nth2[5] // MatrixForm // TeXForm



          same result







          share|improve this answer











          $endgroup$













          • $begingroup$
            thanks.. it work perfectly.
            $endgroup$
            – henry
            5 hours ago










          • $begingroup$
            which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
            $endgroup$
            – henry
            5 hours ago








          • 1




            $begingroup$
            @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
            $endgroup$
            – kglr
            5 hours ago










          • $begingroup$
            final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
            $endgroup$
            – henry
            5 hours ago






          • 1




            $begingroup$
            @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
            $endgroup$
            – kglr
            5 hours ago














          3












          3








          3





          $begingroup$

          Remove the braces around Mod:



          ClearAll[nth]
          
nth[n_] := DiagonalMatrix@Table[a^Mod[k, n], {k, 0, 2 n - 1}]
          

          
nth[5] // MatrixForm // TeXForm



          $left(
          begin{array}{cccccccccc}
          1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & a & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & a^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & a^3 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & a^4 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & a & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & a^2 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^3 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^4 \
          end{array}
          right)$




          Alternatively, you can use SparseArray:



          ClearAll[nth2]
          
nth2[n_] := SparseArray[{k_, k_} :> a^Mod[k, n], {2 n - 1, 2 n - 1}]
          
nth2[5] // MatrixForm // TeXForm



          same result




          or use Band in combination with SparseArray:



          ClearAll[nth3]
          
nth3[n_] := SparseArray[Band[{1, 1}] -> Table[a^Mod[k, n], {k, 0, 2 n - 1}]]
          

          
nth2[5] // MatrixForm // TeXForm



          same result







          share|improve this answer











          $endgroup$



          Remove the braces around Mod:



          ClearAll[nth]
          
nth[n_] := DiagonalMatrix@Table[a^Mod[k, n], {k, 0, 2 n - 1}]
          

          
nth[5] // MatrixForm // TeXForm



          $left(
          begin{array}{cccccccccc}
          1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & a & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & a^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & a^3 & 0 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & a^4 & 0 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & a & 0 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & a^2 & 0 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^3 & 0 \
          0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a^4 \
          end{array}
          right)$




          Alternatively, you can use SparseArray:



          ClearAll[nth2]
          
nth2[n_] := SparseArray[{k_, k_} :> a^Mod[k, n], {2 n - 1, 2 n - 1}]
          
nth2[5] // MatrixForm // TeXForm



          same result




          or use Band in combination with SparseArray:



          ClearAll[nth3]
          
nth3[n_] := SparseArray[Band[{1, 1}] -> Table[a^Mod[k, n], {k, 0, 2 n - 1}]]
          

          
nth2[5] // MatrixForm // TeXForm



          same result








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 5 hours ago

























          answered 6 hours ago









          kglrkglr

          183k10201416




          183k10201416












          • $begingroup$
            thanks.. it work perfectly.
            $endgroup$
            – henry
            5 hours ago










          • $begingroup$
            which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
            $endgroup$
            – henry
            5 hours ago








          • 1




            $begingroup$
            @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
            $endgroup$
            – kglr
            5 hours ago










          • $begingroup$
            final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
            $endgroup$
            – henry
            5 hours ago






          • 1




            $begingroup$
            @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
            $endgroup$
            – kglr
            5 hours ago


















          • $begingroup$
            thanks.. it work perfectly.
            $endgroup$
            – henry
            5 hours ago










          • $begingroup$
            which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
            $endgroup$
            – henry
            5 hours ago








          • 1




            $begingroup$
            @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
            $endgroup$
            – kglr
            5 hours ago










          • $begingroup$
            final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
            $endgroup$
            – henry
            5 hours ago






          • 1




            $begingroup$
            @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
            $endgroup$
            – kglr
            5 hours ago
















          $begingroup$
          thanks.. it work perfectly.
          $endgroup$
          – henry
          5 hours ago




          $begingroup$
          thanks.. it work perfectly.
          $endgroup$
          – henry
          5 hours ago












          $begingroup$
          which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
          $endgroup$
          – henry
          5 hours ago






          $begingroup$
          which one more practical? I mean later i need to compute power of modified verson of this matrix. so it will be much more computation.
          $endgroup$
          – henry
          5 hours ago






          1




          1




          $begingroup$
          @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
          $endgroup$
          – kglr
          5 hours ago




          $begingroup$
          @henry, SparseArray +Band is likely to be the faster than DiagonalMatrix.
          $endgroup$
          – kglr
          5 hours ago












          $begingroup$
          final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
          $endgroup$
          – henry
          5 hours ago




          $begingroup$
          final question. If i use MatrixForm as a part of the function. does have any effect of efficiency of the computation?
          $endgroup$
          – henry
          5 hours ago




          1




          1




          $begingroup$
          @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
          $endgroup$
          – kglr
          5 hours ago




          $begingroup$
          @henry, you should use MatrixForm is only for displaying (MatrixForm acts as a "wrapper", which affects printing, but not evaluation. )
          $endgroup$
          – kglr
          5 hours ago


















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          Алькесар

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          Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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