Non-trivial intersections of row spaces in Matlab












0












$begingroup$


Let $W_p$ and $W_f$ be two subspaces (past and future data in matrix form: rows as basis vectors). Let x be a vector that lies in intersection of these two subspaces. Then $∃$ two coefficient vectors $a,b$ such that



$x = a^T W_p = b^T W_f$



I want to find a, b and x, avoiding trivial intersections (i.e. vectors $a$ and $b$ such that $a^T W_p=0 =b^T W_f$)



I was thinking of proceeding in this way:



$
a^T W_p-b^T W_f = 0\
[a^T -b^T] W =0 \
$



with $W =begin{bmatrix}W_p\W_fend{bmatrix}$. Is it possible to compute $a$ and $b$ based on that? maybe with QR factorization of $W$ to find the null space?






linear-algebra matrices vector-spaces intersection-theory







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $W_p$ and $W_f$ be two subspaces (past and future data in matrix form: rows as basis vectors). Let x be a vector that lies in intersection of these two subspaces. Then $∃$ two coefficient vectors $a,b$ such that



    $x = a^T W_p = b^T W_f$



    I want to find a, b and x, avoiding trivial intersections (i.e. vectors $a$ and $b$ such that $a^T W_p=0 =b^T W_f$)



    I was thinking of proceeding in this way:



    $
    a^T W_p-b^T W_f = 0\
    [a^T -b^T] W =0 \
    $



    with $W =begin{bmatrix}W_p\W_fend{bmatrix}$. Is it possible to compute $a$ and $b$ based on that? maybe with QR factorization of $W$ to find the null space?






    linear-algebra matrices vector-spaces intersection-theory







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $W_p$ and $W_f$ be two subspaces (past and future data in matrix form: rows as basis vectors). Let x be a vector that lies in intersection of these two subspaces. Then $∃$ two coefficient vectors $a,b$ such that



      $x = a^T W_p = b^T W_f$



      I want to find a, b and x, avoiding trivial intersections (i.e. vectors $a$ and $b$ such that $a^T W_p=0 =b^T W_f$)



      I was thinking of proceeding in this way:



      $
      a^T W_p-b^T W_f = 0\
      [a^T -b^T] W =0 \
      $



      with $W =begin{bmatrix}W_p\W_fend{bmatrix}$. Is it possible to compute $a$ and $b$ based on that? maybe with QR factorization of $W$ to find the null space?






      linear-algebra matrices vector-spaces intersection-theory







      share|cite|improve this question











      $endgroup$




      Let $W_p$ and $W_f$ be two subspaces (past and future data in matrix form: rows as basis vectors). Let x be a vector that lies in intersection of these two subspaces. Then $∃$ two coefficient vectors $a,b$ such that



      $x = a^T W_p = b^T W_f$



      I want to find a, b and x, avoiding trivial intersections (i.e. vectors $a$ and $b$ such that $a^T W_p=0 =b^T W_f$)



      I was thinking of proceeding in this way:



      $
      a^T W_p-b^T W_f = 0\
      [a^T -b^T] W =0 \
      $



      with $W =begin{bmatrix}W_p\W_fend{bmatrix}$. Is it possible to compute $a$ and $b$ based on that? maybe with QR factorization of $W$ to find the null space?






      linear-algebra matrices vector-spaces intersection-theory




      linear-algebra matrices vector-spaces intersection-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 21 '18 at 9:51







      Betelgeuse

















      asked Dec 20 '18 at 10:43









      BetelgeuseBetelgeuse

      1214




      1214






















          1 Answer
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          0












          $begingroup$

          Taking the transpose on both sides, your equation reads
          $$
          W^T pmatrix{a\ -b} = 0
          $$
          With this in mind, the easiest approach is probably to use the nullspace function in matlab. In particular:



          W = [Wp; Wf];
          
N = null(W');


          If all you need is a non-trivial solution, take the first column of $N$ (that is, N(:,1)) and split it into the vectors a and -b.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
            $endgroup$
            – Betelgeuse
            Dec 21 '18 at 9:48










          • $begingroup$
            If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:23










          • $begingroup$
            Yes, I was wondering how to proceed if you don't have an $x$
            $endgroup$
            – Betelgeuse
            Jan 3 at 13:24










          • $begingroup$
            If you don't have an $x$, then what makes a pair $a,b$ "correct"?
            $endgroup$
            – Omnomnomnom
            Jan 3 at 13:56











          Your Answer





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          active

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          0












          $begingroup$

          Taking the transpose on both sides, your equation reads
          $$
          W^T pmatrix{a\ -b} = 0
          $$
          With this in mind, the easiest approach is probably to use the nullspace function in matlab. In particular:



          W = [Wp; Wf];
          
N = null(W');


          If all you need is a non-trivial solution, take the first column of $N$ (that is, N(:,1)) and split it into the vectors a and -b.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
            $endgroup$
            – Betelgeuse
            Dec 21 '18 at 9:48










          • $begingroup$
            If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:23










          • $begingroup$
            Yes, I was wondering how to proceed if you don't have an $x$
            $endgroup$
            – Betelgeuse
            Jan 3 at 13:24










          • $begingroup$
            If you don't have an $x$, then what makes a pair $a,b$ "correct"?
            $endgroup$
            – Omnomnomnom
            Jan 3 at 13:56
















          0












          $begingroup$

          Taking the transpose on both sides, your equation reads
          $$
          W^T pmatrix{a\ -b} = 0
          $$
          With this in mind, the easiest approach is probably to use the nullspace function in matlab. In particular:



          W = [Wp; Wf];
          
N = null(W');


          If all you need is a non-trivial solution, take the first column of $N$ (that is, N(:,1)) and split it into the vectors a and -b.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
            $endgroup$
            – Betelgeuse
            Dec 21 '18 at 9:48










          • $begingroup$
            If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:23










          • $begingroup$
            Yes, I was wondering how to proceed if you don't have an $x$
            $endgroup$
            – Betelgeuse
            Jan 3 at 13:24










          • $begingroup$
            If you don't have an $x$, then what makes a pair $a,b$ "correct"?
            $endgroup$
            – Omnomnomnom
            Jan 3 at 13:56














          0












          0








          0





          $begingroup$

          Taking the transpose on both sides, your equation reads
          $$
          W^T pmatrix{a\ -b} = 0
          $$
          With this in mind, the easiest approach is probably to use the nullspace function in matlab. In particular:



          W = [Wp; Wf];
          
N = null(W');


          If all you need is a non-trivial solution, take the first column of $N$ (that is, N(:,1)) and split it into the vectors a and -b.






          share|cite|improve this answer









          $endgroup$



          Taking the transpose on both sides, your equation reads
          $$
          W^T pmatrix{a\ -b} = 0
          $$
          With this in mind, the easiest approach is probably to use the nullspace function in matlab. In particular:



          W = [Wp; Wf];
          
N = null(W');


          If all you need is a non-trivial solution, take the first column of $N$ (that is, N(:,1)) and split it into the vectors a and -b.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 20:43









          OmnomnomnomOmnomnomnom

          128k791184




          128k791184












          • $begingroup$
            This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
            $endgroup$
            – Betelgeuse
            Dec 21 '18 at 9:48










          • $begingroup$
            If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:23










          • $begingroup$
            Yes, I was wondering how to proceed if you don't have an $x$
            $endgroup$
            – Betelgeuse
            Jan 3 at 13:24










          • $begingroup$
            If you don't have an $x$, then what makes a pair $a,b$ "correct"?
            $endgroup$
            – Omnomnomnom
            Jan 3 at 13:56


















          • $begingroup$
            This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
            $endgroup$
            – Betelgeuse
            Dec 21 '18 at 9:48










          • $begingroup$
            If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
            $endgroup$
            – Omnomnomnom
            Dec 21 '18 at 17:23










          • $begingroup$
            Yes, I was wondering how to proceed if you don't have an $x$
            $endgroup$
            – Betelgeuse
            Jan 3 at 13:24










          • $begingroup$
            If you don't have an $x$, then what makes a pair $a,b$ "correct"?
            $endgroup$
            – Omnomnomnom
            Jan 3 at 13:56
















          $begingroup$
          This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
          $endgroup$
          – Betelgeuse
          Dec 21 '18 at 9:48




          $begingroup$
          This procedure is correct in principle, but it does not give me the correct sequence of $a,b$ and $x$ unfortunately (I have already the sequence of $x$ so to check the results).
          $endgroup$
          – Betelgeuse
          Dec 21 '18 at 9:48












          $begingroup$
          If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
          $endgroup$
          – Omnomnomnom
          Dec 21 '18 at 17:23




          $begingroup$
          If you already have an $x$, then you're really solving the system $$ W_p^T a = x\ W_f^T b = x\ $$ which is to say $$ pmatrix{W_p^T&0\0&W_f^T} pmatrix{a\b} = pmatrix{x\x} $$
          $endgroup$
          – Omnomnomnom
          Dec 21 '18 at 17:23












          $begingroup$
          Yes, I was wondering how to proceed if you don't have an $x$
          $endgroup$
          – Betelgeuse
          Jan 3 at 13:24




          $begingroup$
          Yes, I was wondering how to proceed if you don't have an $x$
          $endgroup$
          – Betelgeuse
          Jan 3 at 13:24












          $begingroup$
          If you don't have an $x$, then what makes a pair $a,b$ "correct"?
          $endgroup$
          – Omnomnomnom
          Jan 3 at 13:56




          $begingroup$
          If you don't have an $x$, then what makes a pair $a,b$ "correct"?
          $endgroup$
          – Omnomnomnom
          Jan 3 at 13:56


















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