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 21 mins ago









      m_goldberg

      85.9k872196




      85.9k872196










      asked 5 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 5 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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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要 ...
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          Tu-95轟炸機

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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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