Jordan canonical form of strictly upper or lower triangular matrices of even order n and rank n-1.

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












I am trying to find the jordan canonical form of $
A=
left[ beginarraycccc
0 & 0 & 0 & 0 \
3 & 0 & 0 & 0 \
0 & 2 & 0 & 0 \
0 & 0 & 1 & 0 \
endarray right]
$.
In this process I found $0$ is the only eigen value with algebraic multiplicity $4$ and geometric multiplicity $1$ so jordan canonical form should have only one 0-jordan block of order $4$.
But as $A^3 = 0$, we have $x^3$ as minimal polynomial and that suggest the largest $0$-jordan block can be of order $3$.




Please help, where am I wrong?







linear-algebra linear-transformations







share|cite|improve this question



























    up vote
    1
    down vote

    favorite












    I am trying to find the jordan canonical form of $
    A=
    left[ beginarraycccc
    0 & 0 & 0 & 0 \
    3 & 0 & 0 & 0 \
    0 & 2 & 0 & 0 \
    0 & 0 & 1 & 0 \
    endarray right]
    $.
    In this process I found $0$ is the only eigen value with algebraic multiplicity $4$ and geometric multiplicity $1$ so jordan canonical form should have only one 0-jordan block of order $4$.
    But as $A^3 = 0$, we have $x^3$ as minimal polynomial and that suggest the largest $0$-jordan block can be of order $3$.




    Please help, where am I wrong?







    linear-algebra linear-transformations







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am trying to find the jordan canonical form of $
      A=
      left[ beginarraycccc
      0 & 0 & 0 & 0 \
      3 & 0 & 0 & 0 \
      0 & 2 & 0 & 0 \
      0 & 0 & 1 & 0 \
      endarray right]
      $.
      In this process I found $0$ is the only eigen value with algebraic multiplicity $4$ and geometric multiplicity $1$ so jordan canonical form should have only one 0-jordan block of order $4$.
      But as $A^3 = 0$, we have $x^3$ as minimal polynomial and that suggest the largest $0$-jordan block can be of order $3$.




      Please help, where am I wrong?







      linear-algebra linear-transformations







      share|cite|improve this question















      I am trying to find the jordan canonical form of $
      A=
      left[ beginarraycccc
      0 & 0 & 0 & 0 \
      3 & 0 & 0 & 0 \
      0 & 2 & 0 & 0 \
      0 & 0 & 1 & 0 \
      endarray right]
      $.
      In this process I found $0$ is the only eigen value with algebraic multiplicity $4$ and geometric multiplicity $1$ so jordan canonical form should have only one 0-jordan block of order $4$.
      But as $A^3 = 0$, we have $x^3$ as minimal polynomial and that suggest the largest $0$-jordan block can be of order $3$.




      Please help, where am I wrong?







      linear-algebra linear-transformations




      linear-algebra linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 1 at 8:23









      Anonymous I

      8311726




      8311726










      asked Sep 1 at 7:53









      Voly op

      354




      354




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote



          accepted










          octave:4> A
          A =

           0 0 0 0
           3 0 0 0
           0 2 0 0
           0 0 1 0

          octave:5> A^3
          ans =

           0 0 0 0
           0 0 0 0
           0 0 0 0
           6 0 0 0


          Notice at the lower left corner, the entry is non-zero. The minimal polynomial is indeed $x^4$.






          share|cite|improve this answer




















          • Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
            – Voly op
            Sep 1 at 8:10










          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',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2901476%2fjordan-canonical-form-of-strictly-upper-or-lower-triangular-matrices-of-even-ord%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote



          accepted










          octave:4> A
          A =

           0 0 0 0
           3 0 0 0
           0 2 0 0
           0 0 1 0

          octave:5> A^3
          ans =

           0 0 0 0
           0 0 0 0
           0 0 0 0
           6 0 0 0


          Notice at the lower left corner, the entry is non-zero. The minimal polynomial is indeed $x^4$.






          share|cite|improve this answer




















          • Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
            – Voly op
            Sep 1 at 8:10














          up vote
          0
          down vote



          accepted










          octave:4> A
          A =

           0 0 0 0
           3 0 0 0
           0 2 0 0
           0 0 1 0

          octave:5> A^3
          ans =

           0 0 0 0
           0 0 0 0
           0 0 0 0
           6 0 0 0


          Notice at the lower left corner, the entry is non-zero. The minimal polynomial is indeed $x^4$.






          share|cite|improve this answer




















          • Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
            – Voly op
            Sep 1 at 8:10












          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          octave:4> A
          A =

           0 0 0 0
           3 0 0 0
           0 2 0 0
           0 0 1 0

          octave:5> A^3
          ans =

           0 0 0 0
           0 0 0 0
           0 0 0 0
           6 0 0 0


          Notice at the lower left corner, the entry is non-zero. The minimal polynomial is indeed $x^4$.






          share|cite|improve this answer












          octave:4> A
          A =

           0 0 0 0
           3 0 0 0
           0 2 0 0
           0 0 1 0

          octave:5> A^3
          ans =

           0 0 0 0
           0 0 0 0
           0 0 0 0
           6 0 0 0


          Notice at the lower left corner, the entry is non-zero. The minimal polynomial is indeed $x^4$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 1 at 8:01









          Siong Thye Goh

          81.9k1456104




          81.9k1456104











          • Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
            – Voly op
            Sep 1 at 8:10
















          • Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
            – Voly op
            Sep 1 at 8:10















          Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
          – Voly op
          Sep 1 at 8:10




          Ohh .. I calculated A^4 directly instead of A^3 and thought it was A^3. Poor me. Thank you.
          – Voly op
          Sep 1 at 8:10

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901476%2fjordan-canonical-form-of-strictly-upper-or-lower-triangular-matrices-of-even-ord%23new-answer', 'question_page');

          );

          Post as a guest
































































          -

          這個網誌中的熱門文章

          How to combine Bézier curves to a surface?

          圖片
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
          閱讀完整內容

          Mutual Information Always Non-negative

          圖片
          Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
          閱讀完整內容

          Why am i infinitely getting the same tweet with the Twitter Search API?

          圖片
          up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
          閱讀完整內容