Procedure to find a pair of matrices whose product has rank less than each matrix in the pair












1












$begingroup$


I am trying to figure out how to find two rank-deficient matrices (not necessarily square) which when multiplied will have rank less than either of the original matrices.



In other words I am looking for:



$$A in R^{a times b} , B in R^{c times d}: Rank(AB) < min(Rank(A),Rank(B))$$



Ideally, I would be interested in a procedure which could be used to generate such matrices. Examples or terminology I could search to learn more about the topic would also be very useful.



Appreciate your time and help.






linear-algebra matrices







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am trying to figure out how to find two rank-deficient matrices (not necessarily square) which when multiplied will have rank less than either of the original matrices.



    In other words I am looking for:



    $$A in R^{a times b} , B in R^{c times d}: Rank(AB) < min(Rank(A),Rank(B))$$



    Ideally, I would be interested in a procedure which could be used to generate such matrices. Examples or terminology I could search to learn more about the topic would also be very useful.



    Appreciate your time and help.






    linear-algebra matrices







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to figure out how to find two rank-deficient matrices (not necessarily square) which when multiplied will have rank less than either of the original matrices.



      In other words I am looking for:



      $$A in R^{a times b} , B in R^{c times d}: Rank(AB) < min(Rank(A),Rank(B))$$



      Ideally, I would be interested in a procedure which could be used to generate such matrices. Examples or terminology I could search to learn more about the topic would also be very useful.



      Appreciate your time and help.






      linear-algebra matrices







      share|cite|improve this question









      $endgroup$




      I am trying to figure out how to find two rank-deficient matrices (not necessarily square) which when multiplied will have rank less than either of the original matrices.



      In other words I am looking for:



      $$A in R^{a times b} , B in R^{c times d}: Rank(AB) < min(Rank(A),Rank(B))$$



      Ideally, I would be interested in a procedure which could be used to generate such matrices. Examples or terminology I could search to learn more about the topic would also be very useful.



      Appreciate your time and help.






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 20 at 23:22









      FilipFilip

      177210




      177210






















          2 Answers
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          0












          $begingroup$

          To expand a bit on the idea in José Carlos Santos’ answer, take any matrix $A$ with a nontrivial kernel and rank greater than zero. Let the columns of $B$ be any elements of $ker A$, with at least one nonzero column so that the rank of $B$ is also positive. Then $AB=0$.



          Alternatively, take any set of pairwise orthogonal vectors and split them up however you like among the rows of $A$ and columns of $B$. Once again, $AB=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
            $endgroup$
            – Filip
            Jan 20 at 23:46










          • $begingroup$
            @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
            $endgroup$
            – amd
            Jan 20 at 23:49





















          1












          $begingroup$

          You can take two non-zero matrices $A$ and $B$ such that $AB=0$. For example, take$$A=B=begin{bmatrix}0&1\0&0end{bmatrix}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I believe this is Nilpotency, and this is restricted to square matrices.
            $endgroup$
            – Filip
            Jan 20 at 23:29










          • $begingroup$
            Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
            $endgroup$
            – José Carlos Santos
            Jan 20 at 23:33










          • $begingroup$
            To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
            $endgroup$
            – Klaus
            Jan 20 at 23:38










          • $begingroup$
            So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
            $endgroup$
            – Filip
            Jan 20 at 23:39






          • 2




            $begingroup$
            This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
            $endgroup$
            – Klaus
            Jan 20 at 23:53











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          2 Answers
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          0












          $begingroup$

          To expand a bit on the idea in José Carlos Santos’ answer, take any matrix $A$ with a nontrivial kernel and rank greater than zero. Let the columns of $B$ be any elements of $ker A$, with at least one nonzero column so that the rank of $B$ is also positive. Then $AB=0$.



          Alternatively, take any set of pairwise orthogonal vectors and split them up however you like among the rows of $A$ and columns of $B$. Once again, $AB=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
            $endgroup$
            – Filip
            Jan 20 at 23:46










          • $begingroup$
            @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
            $endgroup$
            – amd
            Jan 20 at 23:49


















          0












          $begingroup$

          To expand a bit on the idea in José Carlos Santos’ answer, take any matrix $A$ with a nontrivial kernel and rank greater than zero. Let the columns of $B$ be any elements of $ker A$, with at least one nonzero column so that the rank of $B$ is also positive. Then $AB=0$.



          Alternatively, take any set of pairwise orthogonal vectors and split them up however you like among the rows of $A$ and columns of $B$. Once again, $AB=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
            $endgroup$
            – Filip
            Jan 20 at 23:46










          • $begingroup$
            @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
            $endgroup$
            – amd
            Jan 20 at 23:49
















          0












          0








          0





          $begingroup$

          To expand a bit on the idea in José Carlos Santos’ answer, take any matrix $A$ with a nontrivial kernel and rank greater than zero. Let the columns of $B$ be any elements of $ker A$, with at least one nonzero column so that the rank of $B$ is also positive. Then $AB=0$.



          Alternatively, take any set of pairwise orthogonal vectors and split them up however you like among the rows of $A$ and columns of $B$. Once again, $AB=0$.






          share|cite|improve this answer









          $endgroup$



          To expand a bit on the idea in José Carlos Santos’ answer, take any matrix $A$ with a nontrivial kernel and rank greater than zero. Let the columns of $B$ be any elements of $ker A$, with at least one nonzero column so that the rank of $B$ is also positive. Then $AB=0$.



          Alternatively, take any set of pairwise orthogonal vectors and split them up however you like among the rows of $A$ and columns of $B$. Once again, $AB=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 23:38









          amdamd

          30.5k21050




          30.5k21050












          • $begingroup$
            This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
            $endgroup$
            – Filip
            Jan 20 at 23:46










          • $begingroup$
            @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
            $endgroup$
            – amd
            Jan 20 at 23:49




















          • $begingroup$
            This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
            $endgroup$
            – Filip
            Jan 20 at 23:46










          • $begingroup$
            @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
            $endgroup$
            – amd
            Jan 20 at 23:49


















          $begingroup$
          This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
          $endgroup$
          – Filip
          Jan 20 at 23:46




          $begingroup$
          This was definitely helpful, I'm trying to think of how I can use your second statement so that AB is not necessarily the zero matrix.
          $endgroup$
          – Filip
          Jan 20 at 23:46












          $begingroup$
          @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
          $endgroup$
          – amd
          Jan 20 at 23:49






          $begingroup$
          @Filip Place one of the nonzero vectors into both matrices. $AB$ then has rank one.
          $endgroup$
          – amd
          Jan 20 at 23:49













          1












          $begingroup$

          You can take two non-zero matrices $A$ and $B$ such that $AB=0$. For example, take$$A=B=begin{bmatrix}0&1\0&0end{bmatrix}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I believe this is Nilpotency, and this is restricted to square matrices.
            $endgroup$
            – Filip
            Jan 20 at 23:29










          • $begingroup$
            Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
            $endgroup$
            – José Carlos Santos
            Jan 20 at 23:33










          • $begingroup$
            To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
            $endgroup$
            – Klaus
            Jan 20 at 23:38










          • $begingroup$
            So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
            $endgroup$
            – Filip
            Jan 20 at 23:39






          • 2




            $begingroup$
            This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
            $endgroup$
            – Klaus
            Jan 20 at 23:53
















          1












          $begingroup$

          You can take two non-zero matrices $A$ and $B$ such that $AB=0$. For example, take$$A=B=begin{bmatrix}0&1\0&0end{bmatrix}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I believe this is Nilpotency, and this is restricted to square matrices.
            $endgroup$
            – Filip
            Jan 20 at 23:29










          • $begingroup$
            Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
            $endgroup$
            – José Carlos Santos
            Jan 20 at 23:33










          • $begingroup$
            To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
            $endgroup$
            – Klaus
            Jan 20 at 23:38










          • $begingroup$
            So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
            $endgroup$
            – Filip
            Jan 20 at 23:39






          • 2




            $begingroup$
            This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
            $endgroup$
            – Klaus
            Jan 20 at 23:53














          1












          1








          1





          $begingroup$

          You can take two non-zero matrices $A$ and $B$ such that $AB=0$. For example, take$$A=B=begin{bmatrix}0&1\0&0end{bmatrix}.$$






          share|cite|improve this answer









          $endgroup$



          You can take two non-zero matrices $A$ and $B$ such that $AB=0$. For example, take$$A=B=begin{bmatrix}0&1\0&0end{bmatrix}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 23:29









          José Carlos SantosJosé Carlos Santos

          163k22130233




          163k22130233












          • $begingroup$
            I believe this is Nilpotency, and this is restricted to square matrices.
            $endgroup$
            – Filip
            Jan 20 at 23:29










          • $begingroup$
            Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
            $endgroup$
            – José Carlos Santos
            Jan 20 at 23:33










          • $begingroup$
            To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
            $endgroup$
            – Klaus
            Jan 20 at 23:38










          • $begingroup$
            So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
            $endgroup$
            – Filip
            Jan 20 at 23:39






          • 2




            $begingroup$
            This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
            $endgroup$
            – Klaus
            Jan 20 at 23:53


















          • $begingroup$
            I believe this is Nilpotency, and this is restricted to square matrices.
            $endgroup$
            – Filip
            Jan 20 at 23:29










          • $begingroup$
            Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
            $endgroup$
            – José Carlos Santos
            Jan 20 at 23:33










          • $begingroup$
            To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
            $endgroup$
            – Klaus
            Jan 20 at 23:38










          • $begingroup$
            So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
            $endgroup$
            – Filip
            Jan 20 at 23:39






          • 2




            $begingroup$
            This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
            $endgroup$
            – Klaus
            Jan 20 at 23:53
















          $begingroup$
          I believe this is Nilpotency, and this is restricted to square matrices.
          $endgroup$
          – Filip
          Jan 20 at 23:29




          $begingroup$
          I believe this is Nilpotency, and this is restricted to square matrices.
          $endgroup$
          – Filip
          Jan 20 at 23:29












          $begingroup$
          Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
          $endgroup$
          – José Carlos Santos
          Jan 20 at 23:33




          $begingroup$
          Not at all. You can also take$$A=begin{bmatrix}1&0end{bmatrix}text{ and }B=begin{bmatrix}0\1end{bmatrix}.$$
          $endgroup$
          – José Carlos Santos
          Jan 20 at 23:33












          $begingroup$
          To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
          $endgroup$
          – Klaus
          Jan 20 at 23:38




          $begingroup$
          To generalize this, you can take arbitrarily many orthogonal vectors and distribute them between $A$ and $B$ as rows or columns, respectively. Fill up with zeros and there you go.
          $endgroup$
          – Klaus
          Jan 20 at 23:38












          $begingroup$
          So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
          $endgroup$
          – Filip
          Jan 20 at 23:39




          $begingroup$
          So in general, if the columns of B belong to the nullspace of A then AB = 0, but I have also been trying to think of an example where the result matrix is not the zero matrix. For example if $$A in R^{4 times 3}, B in R^{3 times 5}, Rank(A)=2, Rank(B)=3: Rank(AB)=1$$
          $endgroup$
          – Filip
          Jan 20 at 23:39




          2




          2




          $begingroup$
          This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
          $endgroup$
          – Klaus
          Jan 20 at 23:53




          $begingroup$
          This is actually impossible. If $B$ has Rank 3, it is surjective. As the rank of $A$ is $2$, $AB$ needs to have at least a two-dimensional range in this configuration.
          $endgroup$
          – Klaus
          Jan 20 at 23:53


















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