Rank of a matrix product of two vectors












1












$begingroup$


Let $lambda=(lambda_1, lambda_2,...,lambda_n)in Bbb R^n$ a vector no zero. Let $A=(a_{ij})$ the matrix of $ntimes n$ so that $a_{ij}=lambda_i lambda_j$. Determine the rank of the matrix $A$.



The matrix $A$ is $$ left(
begin{array}{ccccc}
lambda_1^2 & lambda_1lambda_2 & cdots & lambda_1lambda_n \
lambda_2lambda_1 & lambda_2^2 &cdots & lambda_2lambda_n \
vdots& vdots & ddots & vdots \
lambda_nlambda_1 & lambda_nlambda_2& cdots & lambda_n^2 \
end{array}
right)$$






matrices







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$endgroup$

















    1












    $begingroup$


    Let $lambda=(lambda_1, lambda_2,...,lambda_n)in Bbb R^n$ a vector no zero. Let $A=(a_{ij})$ the matrix of $ntimes n$ so that $a_{ij}=lambda_i lambda_j$. Determine the rank of the matrix $A$.



    The matrix $A$ is $$ left(
    begin{array}{ccccc}
    lambda_1^2 & lambda_1lambda_2 & cdots & lambda_1lambda_n \
    lambda_2lambda_1 & lambda_2^2 &cdots & lambda_2lambda_n \
    vdots& vdots & ddots & vdots \
    lambda_nlambda_1 & lambda_nlambda_2& cdots & lambda_n^2 \
    end{array}
    right)$$






    matrices







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    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $lambda=(lambda_1, lambda_2,...,lambda_n)in Bbb R^n$ a vector no zero. Let $A=(a_{ij})$ the matrix of $ntimes n$ so that $a_{ij}=lambda_i lambda_j$. Determine the rank of the matrix $A$.



      The matrix $A$ is $$ left(
      begin{array}{ccccc}
      lambda_1^2 & lambda_1lambda_2 & cdots & lambda_1lambda_n \
      lambda_2lambda_1 & lambda_2^2 &cdots & lambda_2lambda_n \
      vdots& vdots & ddots & vdots \
      lambda_nlambda_1 & lambda_nlambda_2& cdots & lambda_n^2 \
      end{array}
      right)$$






      matrices







      share|cite|improve this question











      $endgroup$




      Let $lambda=(lambda_1, lambda_2,...,lambda_n)in Bbb R^n$ a vector no zero. Let $A=(a_{ij})$ the matrix of $ntimes n$ so that $a_{ij}=lambda_i lambda_j$. Determine the rank of the matrix $A$.



      The matrix $A$ is $$ left(
      begin{array}{ccccc}
      lambda_1^2 & lambda_1lambda_2 & cdots & lambda_1lambda_n \
      lambda_2lambda_1 & lambda_2^2 &cdots & lambda_2lambda_n \
      vdots& vdots & ddots & vdots \
      lambda_nlambda_1 & lambda_nlambda_2& cdots & lambda_n^2 \
      end{array}
      right)$$






      matrices




      matrices






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      edited Feb 20 '16 at 23:23









      John B

      12.2k51840




      12.2k51840










      asked Jan 10 '15 at 4:35









      Roiner Segura CuberoRoiner Segura Cubero

      1,9151730




      1,9151730






















          4 Answers
          4






          active

          oldest

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          2












          $begingroup$

          Fact: $text{Rank}(AB) le text{min}{text{Rank}(A),text{Rank}(B)}$



          And your matrix is



          $$left[ begin{matrix}lambda_1 \ lambda_2 \ vdots \ lambda_nend{matrix}right]left[ begin{matrix}lambda_1 & lambda_2 & cdots & lambda_nend{matrix}right]$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Hint. Every row of $A$ is a scalar multiple of the row vector $lambda$, and $Ane0$. So, what is the row rank of $A$? Now $A$ is a square matrix. Therefore its rank is equal to its row rank.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              which is the rank of a vector?
              $endgroup$
              – Roiner Segura Cubero
              Jan 10 '15 at 4:46










            • $begingroup$
              @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
              $endgroup$
              – janmarqz
              Jan 10 '15 at 4:50



















            1












            $begingroup$

            The rank is one because you have in the rows scalar multiples of the original $lambda$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do not understand that the rank of the matrix is one
              $endgroup$
              – Roiner Segura Cubero
              Jan 10 '15 at 4:44










            • $begingroup$
              how do you define rank of a matrix?
              $endgroup$
              – janmarqz
              Jan 10 '15 at 4:46



















            0












            $begingroup$

            You may alternatively exploit that the rank of a matrix is preserved under a change of a basis, which is described by similarity transformations.



            Choose an orthogonal transformation $S$, thus $Scirc S^T=mathbb 1$, such that
            $$Sbig(underbrace{(1,dots,0)}_{quad =,e_1}big) :=: frac 1{|lambda|}(lambda_1, lambda_2,dots,lambda_n),.$$
            Now consider $A$ and the matrix products
            $$A:=:big(lambda_1,dots,lambda_nbig)
            begin{pmatrix}lambda_1\ vdots\ lambda_nend{pmatrix}
            ,=, |lambda|,Se_1,big(|lambda|,Se_1big)^T
            ,=, |lambda|^2,S:e_1e_1^T,S^T \[3ex]
            ,=, |lambda|^2,S,
            left(begin{smallmatrix} 1&0&dots&0\ 0&0&&vdots\ vdots&&ddots\
            0&ldots&&0end{smallmatrix}right),S^{-1}qquad
            $$
            from which $A$'s rank can be read off.






            share|cite|improve this answer









            $endgroup$













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              4 Answers
              4






              active

              oldest

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              4 Answers
              4






              active

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              active

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              active

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              2












              $begingroup$

              Fact: $text{Rank}(AB) le text{min}{text{Rank}(A),text{Rank}(B)}$



              And your matrix is



              $$left[ begin{matrix}lambda_1 \ lambda_2 \ vdots \ lambda_nend{matrix}right]left[ begin{matrix}lambda_1 & lambda_2 & cdots & lambda_nend{matrix}right]$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Fact: $text{Rank}(AB) le text{min}{text{Rank}(A),text{Rank}(B)}$



                And your matrix is



                $$left[ begin{matrix}lambda_1 \ lambda_2 \ vdots \ lambda_nend{matrix}right]left[ begin{matrix}lambda_1 & lambda_2 & cdots & lambda_nend{matrix}right]$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Fact: $text{Rank}(AB) le text{min}{text{Rank}(A),text{Rank}(B)}$



                  And your matrix is



                  $$left[ begin{matrix}lambda_1 \ lambda_2 \ vdots \ lambda_nend{matrix}right]left[ begin{matrix}lambda_1 & lambda_2 & cdots & lambda_nend{matrix}right]$$






                  share|cite|improve this answer









                  $endgroup$



                  Fact: $text{Rank}(AB) le text{min}{text{Rank}(A),text{Rank}(B)}$



                  And your matrix is



                  $$left[ begin{matrix}lambda_1 \ lambda_2 \ vdots \ lambda_nend{matrix}right]left[ begin{matrix}lambda_1 & lambda_2 & cdots & lambda_nend{matrix}right]$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 '15 at 4:46









                  David PetersonDavid Peterson

                  8,87821935




                  8,87821935























                      2












                      $begingroup$

                      Hint. Every row of $A$ is a scalar multiple of the row vector $lambda$, and $Ane0$. So, what is the row rank of $A$? Now $A$ is a square matrix. Therefore its rank is equal to its row rank.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        which is the rank of a vector?
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:46










                      • $begingroup$
                        @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:50
















                      2












                      $begingroup$

                      Hint. Every row of $A$ is a scalar multiple of the row vector $lambda$, and $Ane0$. So, what is the row rank of $A$? Now $A$ is a square matrix. Therefore its rank is equal to its row rank.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        which is the rank of a vector?
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:46










                      • $begingroup$
                        @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:50














                      2












                      2








                      2





                      $begingroup$

                      Hint. Every row of $A$ is a scalar multiple of the row vector $lambda$, and $Ane0$. So, what is the row rank of $A$? Now $A$ is a square matrix. Therefore its rank is equal to its row rank.






                      share|cite|improve this answer









                      $endgroup$



                      Hint. Every row of $A$ is a scalar multiple of the row vector $lambda$, and $Ane0$. So, what is the row rank of $A$? Now $A$ is a square matrix. Therefore its rank is equal to its row rank.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 10 '15 at 4:40









                      user1551user1551

                      73.6k566129




                      73.6k566129












                      • $begingroup$
                        which is the rank of a vector?
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:46










                      • $begingroup$
                        @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:50


















                      • $begingroup$
                        which is the rank of a vector?
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:46










                      • $begingroup$
                        @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:50
















                      $begingroup$
                      which is the rank of a vector?
                      $endgroup$
                      – Roiner Segura Cubero
                      Jan 10 '15 at 4:46




                      $begingroup$
                      which is the rank of a vector?
                      $endgroup$
                      – Roiner Segura Cubero
                      Jan 10 '15 at 4:46












                      $begingroup$
                      @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
                      $endgroup$
                      – janmarqz
                      Jan 10 '15 at 4:50




                      $begingroup$
                      @RoinerSeguraCubero: a vector is a matrix $ntimes 1$ or $1times n$
                      $endgroup$
                      – janmarqz
                      Jan 10 '15 at 4:50











                      1












                      $begingroup$

                      The rank is one because you have in the rows scalar multiples of the original $lambda$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I do not understand that the rank of the matrix is one
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:44










                      • $begingroup$
                        how do you define rank of a matrix?
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:46
















                      1












                      $begingroup$

                      The rank is one because you have in the rows scalar multiples of the original $lambda$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I do not understand that the rank of the matrix is one
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:44










                      • $begingroup$
                        how do you define rank of a matrix?
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:46














                      1












                      1








                      1





                      $begingroup$

                      The rank is one because you have in the rows scalar multiples of the original $lambda$.






                      share|cite|improve this answer









                      $endgroup$



                      The rank is one because you have in the rows scalar multiples of the original $lambda$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 10 '15 at 4:40









                      janmarqzjanmarqz

                      6,22741630




                      6,22741630












                      • $begingroup$
                        I do not understand that the rank of the matrix is one
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:44










                      • $begingroup$
                        how do you define rank of a matrix?
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:46


















                      • $begingroup$
                        I do not understand that the rank of the matrix is one
                        $endgroup$
                        – Roiner Segura Cubero
                        Jan 10 '15 at 4:44










                      • $begingroup$
                        how do you define rank of a matrix?
                        $endgroup$
                        – janmarqz
                        Jan 10 '15 at 4:46
















                      $begingroup$
                      I do not understand that the rank of the matrix is one
                      $endgroup$
                      – Roiner Segura Cubero
                      Jan 10 '15 at 4:44




                      $begingroup$
                      I do not understand that the rank of the matrix is one
                      $endgroup$
                      – Roiner Segura Cubero
                      Jan 10 '15 at 4:44












                      $begingroup$
                      how do you define rank of a matrix?
                      $endgroup$
                      – janmarqz
                      Jan 10 '15 at 4:46




                      $begingroup$
                      how do you define rank of a matrix?
                      $endgroup$
                      – janmarqz
                      Jan 10 '15 at 4:46











                      0












                      $begingroup$

                      You may alternatively exploit that the rank of a matrix is preserved under a change of a basis, which is described by similarity transformations.



                      Choose an orthogonal transformation $S$, thus $Scirc S^T=mathbb 1$, such that
                      $$Sbig(underbrace{(1,dots,0)}_{quad =,e_1}big) :=: frac 1{|lambda|}(lambda_1, lambda_2,dots,lambda_n),.$$
                      Now consider $A$ and the matrix products
                      $$A:=:big(lambda_1,dots,lambda_nbig)
                      begin{pmatrix}lambda_1\ vdots\ lambda_nend{pmatrix}
                      ,=, |lambda|,Se_1,big(|lambda|,Se_1big)^T
                      ,=, |lambda|^2,S:e_1e_1^T,S^T \[3ex]
                      ,=, |lambda|^2,S,
                      left(begin{smallmatrix} 1&0&dots&0\ 0&0&&vdots\ vdots&&ddots\
                      0&ldots&&0end{smallmatrix}right),S^{-1}qquad
                      $$
                      from which $A$'s rank can be read off.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        You may alternatively exploit that the rank of a matrix is preserved under a change of a basis, which is described by similarity transformations.



                        Choose an orthogonal transformation $S$, thus $Scirc S^T=mathbb 1$, such that
                        $$Sbig(underbrace{(1,dots,0)}_{quad =,e_1}big) :=: frac 1{|lambda|}(lambda_1, lambda_2,dots,lambda_n),.$$
                        Now consider $A$ and the matrix products
                        $$A:=:big(lambda_1,dots,lambda_nbig)
                        begin{pmatrix}lambda_1\ vdots\ lambda_nend{pmatrix}
                        ,=, |lambda|,Se_1,big(|lambda|,Se_1big)^T
                        ,=, |lambda|^2,S:e_1e_1^T,S^T \[3ex]
                        ,=, |lambda|^2,S,
                        left(begin{smallmatrix} 1&0&dots&0\ 0&0&&vdots\ vdots&&ddots\
                        0&ldots&&0end{smallmatrix}right),S^{-1}qquad
                        $$
                        from which $A$'s rank can be read off.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          You may alternatively exploit that the rank of a matrix is preserved under a change of a basis, which is described by similarity transformations.



                          Choose an orthogonal transformation $S$, thus $Scirc S^T=mathbb 1$, such that
                          $$Sbig(underbrace{(1,dots,0)}_{quad =,e_1}big) :=: frac 1{|lambda|}(lambda_1, lambda_2,dots,lambda_n),.$$
                          Now consider $A$ and the matrix products
                          $$A:=:big(lambda_1,dots,lambda_nbig)
                          begin{pmatrix}lambda_1\ vdots\ lambda_nend{pmatrix}
                          ,=, |lambda|,Se_1,big(|lambda|,Se_1big)^T
                          ,=, |lambda|^2,S:e_1e_1^T,S^T \[3ex]
                          ,=, |lambda|^2,S,
                          left(begin{smallmatrix} 1&0&dots&0\ 0&0&&vdots\ vdots&&ddots\
                          0&ldots&&0end{smallmatrix}right),S^{-1}qquad
                          $$
                          from which $A$'s rank can be read off.






                          share|cite|improve this answer









                          $endgroup$



                          You may alternatively exploit that the rank of a matrix is preserved under a change of a basis, which is described by similarity transformations.



                          Choose an orthogonal transformation $S$, thus $Scirc S^T=mathbb 1$, such that
                          $$Sbig(underbrace{(1,dots,0)}_{quad =,e_1}big) :=: frac 1{|lambda|}(lambda_1, lambda_2,dots,lambda_n),.$$
                          Now consider $A$ and the matrix products
                          $$A:=:big(lambda_1,dots,lambda_nbig)
                          begin{pmatrix}lambda_1\ vdots\ lambda_nend{pmatrix}
                          ,=, |lambda|,Se_1,big(|lambda|,Se_1big)^T
                          ,=, |lambda|^2,S:e_1e_1^T,S^T \[3ex]
                          ,=, |lambda|^2,S,
                          left(begin{smallmatrix} 1&0&dots&0\ 0&0&&vdots\ vdots&&ddots\
                          0&ldots&&0end{smallmatrix}right),S^{-1}qquad
                          $$
                          from which $A$'s rank can be read off.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 4 at 1:27









                          HannoHanno

                          2,284628




                          2,284628






























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