Does there exist a “stochastically universal” two-column matrix?












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$begingroup$


Let $mathcal{R}_{mtimes n}$ denote the family of all $mtimes n$ matrices with elements in $[0,1]$, and let $mathcal{S}_{mtimes n}$ denote the family of all $mtimes n$ row-stochastic matrices, that is, matrices $mathbf{A}inmathcal{R}_{mtimes n}$ with sum of elements in each row equal to $1$.
Let us say that a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if for every $mge 1$ and every matrix $mathbf{B}inmathcal{R}_{mtimes 2}$ there exists a row-stochastic matrix $mathbf{C}inmathcal{S}_{mtimes n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.



Question: Does there exist a stochastically universal matrix $mathbf{A}inmathcal{R}_{ntimes 2}$, for some $nge 1$? If so, how to characterize all stochastically universal matrices?



Note: This question is motivated by an attempt to answer a question on probability kernels. When considering finite measurable spaces, probability kernels correspond to stochastic matrices and the composition of kernels corresponds to the matrix multiplication. A matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ encodes two stochastic matrices in $mathcal{S}_{ntimes 2}$, that is, two probability kernels from an $n$-element measurable space into two $2$-element measurable spaces.






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    $begingroup$


    Let $mathcal{R}_{mtimes n}$ denote the family of all $mtimes n$ matrices with elements in $[0,1]$, and let $mathcal{S}_{mtimes n}$ denote the family of all $mtimes n$ row-stochastic matrices, that is, matrices $mathbf{A}inmathcal{R}_{mtimes n}$ with sum of elements in each row equal to $1$.
    Let us say that a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if for every $mge 1$ and every matrix $mathbf{B}inmathcal{R}_{mtimes 2}$ there exists a row-stochastic matrix $mathbf{C}inmathcal{S}_{mtimes n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.



    Question: Does there exist a stochastically universal matrix $mathbf{A}inmathcal{R}_{ntimes 2}$, for some $nge 1$? If so, how to characterize all stochastically universal matrices?



    Note: This question is motivated by an attempt to answer a question on probability kernels. When considering finite measurable spaces, probability kernels correspond to stochastic matrices and the composition of kernels corresponds to the matrix multiplication. A matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ encodes two stochastic matrices in $mathcal{S}_{ntimes 2}$, that is, two probability kernels from an $n$-element measurable space into two $2$-element measurable spaces.






    linear-algebra matrices







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    $endgroup$















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      $begingroup$


      Let $mathcal{R}_{mtimes n}$ denote the family of all $mtimes n$ matrices with elements in $[0,1]$, and let $mathcal{S}_{mtimes n}$ denote the family of all $mtimes n$ row-stochastic matrices, that is, matrices $mathbf{A}inmathcal{R}_{mtimes n}$ with sum of elements in each row equal to $1$.
      Let us say that a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if for every $mge 1$ and every matrix $mathbf{B}inmathcal{R}_{mtimes 2}$ there exists a row-stochastic matrix $mathbf{C}inmathcal{S}_{mtimes n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.



      Question: Does there exist a stochastically universal matrix $mathbf{A}inmathcal{R}_{ntimes 2}$, for some $nge 1$? If so, how to characterize all stochastically universal matrices?



      Note: This question is motivated by an attempt to answer a question on probability kernels. When considering finite measurable spaces, probability kernels correspond to stochastic matrices and the composition of kernels corresponds to the matrix multiplication. A matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ encodes two stochastic matrices in $mathcal{S}_{ntimes 2}$, that is, two probability kernels from an $n$-element measurable space into two $2$-element measurable spaces.






      linear-algebra matrices







      share|cite|improve this question











      $endgroup$




      Let $mathcal{R}_{mtimes n}$ denote the family of all $mtimes n$ matrices with elements in $[0,1]$, and let $mathcal{S}_{mtimes n}$ denote the family of all $mtimes n$ row-stochastic matrices, that is, matrices $mathbf{A}inmathcal{R}_{mtimes n}$ with sum of elements in each row equal to $1$.
      Let us say that a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if for every $mge 1$ and every matrix $mathbf{B}inmathcal{R}_{mtimes 2}$ there exists a row-stochastic matrix $mathbf{C}inmathcal{S}_{mtimes n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.



      Question: Does there exist a stochastically universal matrix $mathbf{A}inmathcal{R}_{ntimes 2}$, for some $nge 1$? If so, how to characterize all stochastically universal matrices?



      Note: This question is motivated by an attempt to answer a question on probability kernels. When considering finite measurable spaces, probability kernels correspond to stochastic matrices and the composition of kernels corresponds to the matrix multiplication. A matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ encodes two stochastic matrices in $mathcal{S}_{ntimes 2}$, that is, two probability kernels from an $n$-element measurable space into two $2$-element measurable spaces.






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      linear-algebra matrices






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      edited Jan 23 at 9:17







      Peter Elias

















      asked Jan 22 at 10:58









      Peter EliasPeter Elias

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          Let $mathbf{A}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$.
          Then for every $mathbf{B}=begin{pmatrix}a_1&b_1\a_2&b_2\vdots&vdots\a_m&b_mend{pmatrix}inmathcal{R}_{mtimes 2}$ we can take
          $$mathbf{C}=begin{pmatrix}a_1-u_1&b_1-u_1&u_1&1-a_1-b_1+u_1\a_2-u_2&b_2-u_2&u_2&1-a_2-b_2+u_2\vdots&vdots&vdots&vdots\a_m-u_m&b_m-u_m&u_m&1-a_m-b_m+u_mend{pmatrix},$$ where $u_i=min{a_i,b_i}$, for every $iin{1,2,dots,m}$. It is straightforward to check that $mathbf{C}$ is a row-stochastic matrix and $mathbf{B}=mathbf{C}mathbf{A}$, hence $mathbf{A}$ is stochastically universal.



          It is also clear that if $mathbf{A}inmathcal{R}_{ntimes 2}$ contains rows $(1,0)$, $(0,1)$, $(1,1)$, $(0,0)$ then $mathbf{A}$ is stochastically universal.



          Let us show that this condition is also necessary.
          Let $mathbf{A}inmathcal{R}_{ntimes 2}$ be stochastically universal, and let $mathbf{B}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$. By the universality of $mathbf{A}$ there exists $mathbf{C}inmathcal{S}_{4times n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.
          We prove that for every $iin{1,dots,4}$ there exists $jin{1,dots,n}$ such that $b_{i,1}=a_{j,1}$ and $b_{i,2}=a_{j,2}$. Let $iin{1,dots,4}$ and $kin{1,2}$ be fixed. For $jin{1,dots,n}$ denote $a'_{j,k}=left|b_{i,k}-a_{j,k}right|$. Then either $a'_{j,k}=a_{j,k}-b_{i,k}$ for all $j$, or $a'_{j,k}=b_{i,k}-a_{j,k}$ for all $j$, hence $$sum_{j=1}^nc_{i,j},a'_{j,k}=left|b_{i,k}-sum_{j=1}^nc_{i,j},a_{j,k}right|=0.$$
          Since $sum_{j=1}^n c_{i,j}=1$ and $c_{i,j}ge 0$ for every $j$, there exists $j$ such that $c_{i,j}>0$. Then $a'_{j,1}=a'_{j,2}=0$, hence $a_{j,1}=b_{i,1}$ and $a_{j,2}=b_{i,2}$.



          Thus, a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if and only if it contains rows $(1,0)$, $(0,1)$, $(1,1)$ and $(0,0)$.






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            $begingroup$

            Let $mathbf{A}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$.
            Then for every $mathbf{B}=begin{pmatrix}a_1&b_1\a_2&b_2\vdots&vdots\a_m&b_mend{pmatrix}inmathcal{R}_{mtimes 2}$ we can take
            $$mathbf{C}=begin{pmatrix}a_1-u_1&b_1-u_1&u_1&1-a_1-b_1+u_1\a_2-u_2&b_2-u_2&u_2&1-a_2-b_2+u_2\vdots&vdots&vdots&vdots\a_m-u_m&b_m-u_m&u_m&1-a_m-b_m+u_mend{pmatrix},$$ where $u_i=min{a_i,b_i}$, for every $iin{1,2,dots,m}$. It is straightforward to check that $mathbf{C}$ is a row-stochastic matrix and $mathbf{B}=mathbf{C}mathbf{A}$, hence $mathbf{A}$ is stochastically universal.



            It is also clear that if $mathbf{A}inmathcal{R}_{ntimes 2}$ contains rows $(1,0)$, $(0,1)$, $(1,1)$, $(0,0)$ then $mathbf{A}$ is stochastically universal.



            Let us show that this condition is also necessary.
            Let $mathbf{A}inmathcal{R}_{ntimes 2}$ be stochastically universal, and let $mathbf{B}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$. By the universality of $mathbf{A}$ there exists $mathbf{C}inmathcal{S}_{4times n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.
            We prove that for every $iin{1,dots,4}$ there exists $jin{1,dots,n}$ such that $b_{i,1}=a_{j,1}$ and $b_{i,2}=a_{j,2}$. Let $iin{1,dots,4}$ and $kin{1,2}$ be fixed. For $jin{1,dots,n}$ denote $a'_{j,k}=left|b_{i,k}-a_{j,k}right|$. Then either $a'_{j,k}=a_{j,k}-b_{i,k}$ for all $j$, or $a'_{j,k}=b_{i,k}-a_{j,k}$ for all $j$, hence $$sum_{j=1}^nc_{i,j},a'_{j,k}=left|b_{i,k}-sum_{j=1}^nc_{i,j},a_{j,k}right|=0.$$
            Since $sum_{j=1}^n c_{i,j}=1$ and $c_{i,j}ge 0$ for every $j$, there exists $j$ such that $c_{i,j}>0$. Then $a'_{j,1}=a'_{j,2}=0$, hence $a_{j,1}=b_{i,1}$ and $a_{j,2}=b_{i,2}$.



            Thus, a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if and only if it contains rows $(1,0)$, $(0,1)$, $(1,1)$ and $(0,0)$.






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              0












              $begingroup$

              Let $mathbf{A}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$.
              Then for every $mathbf{B}=begin{pmatrix}a_1&b_1\a_2&b_2\vdots&vdots\a_m&b_mend{pmatrix}inmathcal{R}_{mtimes 2}$ we can take
              $$mathbf{C}=begin{pmatrix}a_1-u_1&b_1-u_1&u_1&1-a_1-b_1+u_1\a_2-u_2&b_2-u_2&u_2&1-a_2-b_2+u_2\vdots&vdots&vdots&vdots\a_m-u_m&b_m-u_m&u_m&1-a_m-b_m+u_mend{pmatrix},$$ where $u_i=min{a_i,b_i}$, for every $iin{1,2,dots,m}$. It is straightforward to check that $mathbf{C}$ is a row-stochastic matrix and $mathbf{B}=mathbf{C}mathbf{A}$, hence $mathbf{A}$ is stochastically universal.



              It is also clear that if $mathbf{A}inmathcal{R}_{ntimes 2}$ contains rows $(1,0)$, $(0,1)$, $(1,1)$, $(0,0)$ then $mathbf{A}$ is stochastically universal.



              Let us show that this condition is also necessary.
              Let $mathbf{A}inmathcal{R}_{ntimes 2}$ be stochastically universal, and let $mathbf{B}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$. By the universality of $mathbf{A}$ there exists $mathbf{C}inmathcal{S}_{4times n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.
              We prove that for every $iin{1,dots,4}$ there exists $jin{1,dots,n}$ such that $b_{i,1}=a_{j,1}$ and $b_{i,2}=a_{j,2}$. Let $iin{1,dots,4}$ and $kin{1,2}$ be fixed. For $jin{1,dots,n}$ denote $a'_{j,k}=left|b_{i,k}-a_{j,k}right|$. Then either $a'_{j,k}=a_{j,k}-b_{i,k}$ for all $j$, or $a'_{j,k}=b_{i,k}-a_{j,k}$ for all $j$, hence $$sum_{j=1}^nc_{i,j},a'_{j,k}=left|b_{i,k}-sum_{j=1}^nc_{i,j},a_{j,k}right|=0.$$
              Since $sum_{j=1}^n c_{i,j}=1$ and $c_{i,j}ge 0$ for every $j$, there exists $j$ such that $c_{i,j}>0$. Then $a'_{j,1}=a'_{j,2}=0$, hence $a_{j,1}=b_{i,1}$ and $a_{j,2}=b_{i,2}$.



              Thus, a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if and only if it contains rows $(1,0)$, $(0,1)$, $(1,1)$ and $(0,0)$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $mathbf{A}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$.
                Then for every $mathbf{B}=begin{pmatrix}a_1&b_1\a_2&b_2\vdots&vdots\a_m&b_mend{pmatrix}inmathcal{R}_{mtimes 2}$ we can take
                $$mathbf{C}=begin{pmatrix}a_1-u_1&b_1-u_1&u_1&1-a_1-b_1+u_1\a_2-u_2&b_2-u_2&u_2&1-a_2-b_2+u_2\vdots&vdots&vdots&vdots\a_m-u_m&b_m-u_m&u_m&1-a_m-b_m+u_mend{pmatrix},$$ where $u_i=min{a_i,b_i}$, for every $iin{1,2,dots,m}$. It is straightforward to check that $mathbf{C}$ is a row-stochastic matrix and $mathbf{B}=mathbf{C}mathbf{A}$, hence $mathbf{A}$ is stochastically universal.



                It is also clear that if $mathbf{A}inmathcal{R}_{ntimes 2}$ contains rows $(1,0)$, $(0,1)$, $(1,1)$, $(0,0)$ then $mathbf{A}$ is stochastically universal.



                Let us show that this condition is also necessary.
                Let $mathbf{A}inmathcal{R}_{ntimes 2}$ be stochastically universal, and let $mathbf{B}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$. By the universality of $mathbf{A}$ there exists $mathbf{C}inmathcal{S}_{4times n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.
                We prove that for every $iin{1,dots,4}$ there exists $jin{1,dots,n}$ such that $b_{i,1}=a_{j,1}$ and $b_{i,2}=a_{j,2}$. Let $iin{1,dots,4}$ and $kin{1,2}$ be fixed. For $jin{1,dots,n}$ denote $a'_{j,k}=left|b_{i,k}-a_{j,k}right|$. Then either $a'_{j,k}=a_{j,k}-b_{i,k}$ for all $j$, or $a'_{j,k}=b_{i,k}-a_{j,k}$ for all $j$, hence $$sum_{j=1}^nc_{i,j},a'_{j,k}=left|b_{i,k}-sum_{j=1}^nc_{i,j},a_{j,k}right|=0.$$
                Since $sum_{j=1}^n c_{i,j}=1$ and $c_{i,j}ge 0$ for every $j$, there exists $j$ such that $c_{i,j}>0$. Then $a'_{j,1}=a'_{j,2}=0$, hence $a_{j,1}=b_{i,1}$ and $a_{j,2}=b_{i,2}$.



                Thus, a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if and only if it contains rows $(1,0)$, $(0,1)$, $(1,1)$ and $(0,0)$.






                share|cite|improve this answer











                $endgroup$



                Let $mathbf{A}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$.
                Then for every $mathbf{B}=begin{pmatrix}a_1&b_1\a_2&b_2\vdots&vdots\a_m&b_mend{pmatrix}inmathcal{R}_{mtimes 2}$ we can take
                $$mathbf{C}=begin{pmatrix}a_1-u_1&b_1-u_1&u_1&1-a_1-b_1+u_1\a_2-u_2&b_2-u_2&u_2&1-a_2-b_2+u_2\vdots&vdots&vdots&vdots\a_m-u_m&b_m-u_m&u_m&1-a_m-b_m+u_mend{pmatrix},$$ where $u_i=min{a_i,b_i}$, for every $iin{1,2,dots,m}$. It is straightforward to check that $mathbf{C}$ is a row-stochastic matrix and $mathbf{B}=mathbf{C}mathbf{A}$, hence $mathbf{A}$ is stochastically universal.



                It is also clear that if $mathbf{A}inmathcal{R}_{ntimes 2}$ contains rows $(1,0)$, $(0,1)$, $(1,1)$, $(0,0)$ then $mathbf{A}$ is stochastically universal.



                Let us show that this condition is also necessary.
                Let $mathbf{A}inmathcal{R}_{ntimes 2}$ be stochastically universal, and let $mathbf{B}=begin{pmatrix}1&0\0&1\1&1\0&0end{pmatrix}$. By the universality of $mathbf{A}$ there exists $mathbf{C}inmathcal{S}_{4times n}$ such that $mathbf{B}=mathbf{C}mathbf{A}$.
                We prove that for every $iin{1,dots,4}$ there exists $jin{1,dots,n}$ such that $b_{i,1}=a_{j,1}$ and $b_{i,2}=a_{j,2}$. Let $iin{1,dots,4}$ and $kin{1,2}$ be fixed. For $jin{1,dots,n}$ denote $a'_{j,k}=left|b_{i,k}-a_{j,k}right|$. Then either $a'_{j,k}=a_{j,k}-b_{i,k}$ for all $j$, or $a'_{j,k}=b_{i,k}-a_{j,k}$ for all $j$, hence $$sum_{j=1}^nc_{i,j},a'_{j,k}=left|b_{i,k}-sum_{j=1}^nc_{i,j},a_{j,k}right|=0.$$
                Since $sum_{j=1}^n c_{i,j}=1$ and $c_{i,j}ge 0$ for every $j$, there exists $j$ such that $c_{i,j}>0$. Then $a'_{j,1}=a'_{j,2}=0$, hence $a_{j,1}=b_{i,1}$ and $a_{j,2}=b_{i,2}$.



                Thus, a matrix $mathbf{A}inmathcal{R}_{ntimes 2}$ is stochastically universal if and only if it contains rows $(1,0)$, $(0,1)$, $(1,1)$ and $(0,0)$.







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                edited Jan 23 at 9:19

























                answered Jan 22 at 12:24









                Peter EliasPeter Elias

                938415




                938415






























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