Rank of Vandermond form of four matrices

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Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.



Could we say the following matrix (4n*4n)



$beginbmatrix
A&B&C&D\
A&2B&3C&4D
endbmatrix$
always has a full rank?




linear-algebra




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    up vote
    0
    down vote

    favorite












    Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.



    Could we say the following matrix (4n*4n)



    $beginbmatrix
    A&B&C&D\
    A&2B&3C&4D
    endbmatrix$
    always has a full rank?




    linear-algebra




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.



      Could we say the following matrix (4n*4n)



      $beginbmatrix
      A&B&C&D\
      A&2B&3C&4D
      endbmatrix$
      always has a full rank?




      linear-algebra




      share|cite|improve this question













      Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.



      Could we say the following matrix (4n*4n)



      $beginbmatrix
      A&B&C&D\
      A&2B&3C&4D
      endbmatrix$
      always has a full rank?





      linear-algebra





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 4 at 1:13
























      asked Aug 3 at 23:21









      spacehopper

      92




      92




















          1 Answer
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          Here is a counterexample:
          $$
          beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
          $$






          share|cite|improve this answer





















          • Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
            – spacehopper
            Aug 4 at 2:40










          • How could you have then that each 2-combi is full rank (especially when $n=1$)?
            – amsmath
            Aug 4 at 3:18










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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote













          Here is a counterexample:
          $$
          beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
          $$






          share|cite|improve this answer





















          • Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
            – spacehopper
            Aug 4 at 2:40










          • How could you have then that each 2-combi is full rank (especially when $n=1$)?
            – amsmath
            Aug 4 at 3:18














          up vote
          0
          down vote













          Here is a counterexample:
          $$
          beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
          $$






          share|cite|improve this answer





















          • Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
            – spacehopper
            Aug 4 at 2:40










          • How could you have then that each 2-combi is full rank (especially when $n=1$)?
            – amsmath
            Aug 4 at 3:18












          up vote
          0
          down vote










          up vote
          0
          down vote









          Here is a counterexample:
          $$
          beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
          $$






          share|cite|improve this answer













          Here is a counterexample:
          $$
          beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
          $$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 4 at 2:25









          amsmath

          1,545113




          1,545113











          • Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
            – spacehopper
            Aug 4 at 2:40










          • How could you have then that each 2-combi is full rank (especially when $n=1$)?
            – amsmath
            Aug 4 at 3:18
















          • Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
            – spacehopper
            Aug 4 at 2:40










          • How could you have then that each 2-combi is full rank (especially when $n=1$)?
            – amsmath
            Aug 4 at 3:18















          Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
          – spacehopper
          Aug 4 at 2:40




          Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
          – spacehopper
          Aug 4 at 2:40












          How could you have then that each 2-combi is full rank (especially when $n=1$)?
          – amsmath
          Aug 4 at 3:18




          How could you have then that each 2-combi is full rank (especially when $n=1$)?
          – amsmath
          Aug 4 at 3:18












           

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