Is it possible to find matrices $M, N$ such that $MXN$ has specified structure for all $X$?












1














Suppose $M in mathbb R^{2 times *}$ and $N in mathbb R^{# times 2}$. Here $*$ and $#$ means we can pick any dimension we want. The end goal is to find fixed $M, N$ such that range of the linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$. I tried to pick $* = 1, 2$ and $#= 1, 2$ but with no luck. Is this even possible?






linear-algebra matrices







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  • What if $X=0$ and $alpha neq 0$?
    – parsiad
    Nov 22 at 3:29










  • @parsiad: Sorry, this is unclear formulation. I really mean: can we find fixed $M, N$ such that range of linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$.
    – user1101010
    Nov 22 at 3:40


















1














Suppose $M in mathbb R^{2 times *}$ and $N in mathbb R^{# times 2}$. Here $*$ and $#$ means we can pick any dimension we want. The end goal is to find fixed $M, N$ such that range of the linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$. I tried to pick $* = 1, 2$ and $#= 1, 2$ but with no luck. Is this even possible?






linear-algebra matrices







share|cite|improve this question
























  • What if $X=0$ and $alpha neq 0$?
    – parsiad
    Nov 22 at 3:29










  • @parsiad: Sorry, this is unclear formulation. I really mean: can we find fixed $M, N$ such that range of linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$.
    – user1101010
    Nov 22 at 3:40
















1












1








1







Suppose $M in mathbb R^{2 times *}$ and $N in mathbb R^{# times 2}$. Here $*$ and $#$ means we can pick any dimension we want. The end goal is to find fixed $M, N$ such that range of the linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$. I tried to pick $* = 1, 2$ and $#= 1, 2$ but with no luck. Is this even possible?






linear-algebra matrices







share|cite|improve this question















Suppose $M in mathbb R^{2 times *}$ and $N in mathbb R^{# times 2}$. Here $*$ and $#$ means we can pick any dimension we want. The end goal is to find fixed $M, N$ such that range of the linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$. I tried to pick $* = 1, 2$ and $#= 1, 2$ but with no luck. Is this even possible?






linear-algebra matrices




linear-algebra matrices






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edited Nov 22 at 3:41

























asked Nov 22 at 3:26









user1101010

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  • What if $X=0$ and $alpha neq 0$?
    – parsiad
    Nov 22 at 3:29










  • @parsiad: Sorry, this is unclear formulation. I really mean: can we find fixed $M, N$ such that range of linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$.
    – user1101010
    Nov 22 at 3:40




















  • What if $X=0$ and $alpha neq 0$?
    – parsiad
    Nov 22 at 3:29










  • @parsiad: Sorry, this is unclear formulation. I really mean: can we find fixed $M, N$ such that range of linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$.
    – user1101010
    Nov 22 at 3:40


















What if $X=0$ and $alpha neq 0$?
– parsiad
Nov 22 at 3:29




What if $X=0$ and $alpha neq 0$?
– parsiad
Nov 22 at 3:29












@parsiad: Sorry, this is unclear formulation. I really mean: can we find fixed $M, N$ such that range of linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$.
– user1101010
Nov 22 at 3:40






@parsiad: Sorry, this is unclear formulation. I really mean: can we find fixed $M, N$ such that range of linear transformation $T: mathbb R^{* times #} to mathbb R^{2 times 2}$ given by $X mapsto MXN$ is the $1$-dim space spanned by $pmatrix{0 & 1 \ 1 & 0}$.
– user1101010
Nov 22 at 3:40












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No, it's not possible unless you accept that the range is ${0}$. Consider first the case $1,1$. We have
$$
begin{bmatrix} M_{11}\ M_{21}end{bmatrix}begin{bmatrix} N_{11}&N_{12}end{bmatrix}=begin{bmatrix} 0&a\a&0end{bmatrix}.
$$
Then $M_{11}N_{11}=0$. If $M_{11}ne0$, then $N_{11}=0$. But then $a=M_{21}N_{11}=0$. From the $1,2$ entry, $0=a=M_{11}N_{12}$, $N_{12}=0$ and $N=0$. When $N_{11}=0$, we obtain similarly that $M=0$ (just take transpose and apply the argument).



Now in general, if $E_{kj}$ is the matrix with $1$ in the $k,j$ entry and zeroes elsewhere,
$$
ME_{kj}N=M_kN_j,
$$
where $M_k$ is the $k^{rm th}$ column of $M$ and $N_j$ with the $j^{rm th}$ row of $j$. By the previous case, one of them has to be zero. So if $M$ has a nonzero column, then $N=0$; and if $N$ has a nonzero row, then $M=0$.






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    No, it's not possible unless you accept that the range is ${0}$. Consider first the case $1,1$. We have
    $$
    begin{bmatrix} M_{11}\ M_{21}end{bmatrix}begin{bmatrix} N_{11}&N_{12}end{bmatrix}=begin{bmatrix} 0&a\a&0end{bmatrix}.
    $$
    Then $M_{11}N_{11}=0$. If $M_{11}ne0$, then $N_{11}=0$. But then $a=M_{21}N_{11}=0$. From the $1,2$ entry, $0=a=M_{11}N_{12}$, $N_{12}=0$ and $N=0$. When $N_{11}=0$, we obtain similarly that $M=0$ (just take transpose and apply the argument).



    Now in general, if $E_{kj}$ is the matrix with $1$ in the $k,j$ entry and zeroes elsewhere,
    $$
    ME_{kj}N=M_kN_j,
    $$
    where $M_k$ is the $k^{rm th}$ column of $M$ and $N_j$ with the $j^{rm th}$ row of $j$. By the previous case, one of them has to be zero. So if $M$ has a nonzero column, then $N=0$; and if $N$ has a nonzero row, then $M=0$.






    share|cite|improve this answer




























      1














      No, it's not possible unless you accept that the range is ${0}$. Consider first the case $1,1$. We have
      $$
      begin{bmatrix} M_{11}\ M_{21}end{bmatrix}begin{bmatrix} N_{11}&N_{12}end{bmatrix}=begin{bmatrix} 0&a\a&0end{bmatrix}.
      $$
      Then $M_{11}N_{11}=0$. If $M_{11}ne0$, then $N_{11}=0$. But then $a=M_{21}N_{11}=0$. From the $1,2$ entry, $0=a=M_{11}N_{12}$, $N_{12}=0$ and $N=0$. When $N_{11}=0$, we obtain similarly that $M=0$ (just take transpose and apply the argument).



      Now in general, if $E_{kj}$ is the matrix with $1$ in the $k,j$ entry and zeroes elsewhere,
      $$
      ME_{kj}N=M_kN_j,
      $$
      where $M_k$ is the $k^{rm th}$ column of $M$ and $N_j$ with the $j^{rm th}$ row of $j$. By the previous case, one of them has to be zero. So if $M$ has a nonzero column, then $N=0$; and if $N$ has a nonzero row, then $M=0$.






      share|cite|improve this answer


























        1












        1








        1






        No, it's not possible unless you accept that the range is ${0}$. Consider first the case $1,1$. We have
        $$
        begin{bmatrix} M_{11}\ M_{21}end{bmatrix}begin{bmatrix} N_{11}&N_{12}end{bmatrix}=begin{bmatrix} 0&a\a&0end{bmatrix}.
        $$
        Then $M_{11}N_{11}=0$. If $M_{11}ne0$, then $N_{11}=0$. But then $a=M_{21}N_{11}=0$. From the $1,2$ entry, $0=a=M_{11}N_{12}$, $N_{12}=0$ and $N=0$. When $N_{11}=0$, we obtain similarly that $M=0$ (just take transpose and apply the argument).



        Now in general, if $E_{kj}$ is the matrix with $1$ in the $k,j$ entry and zeroes elsewhere,
        $$
        ME_{kj}N=M_kN_j,
        $$
        where $M_k$ is the $k^{rm th}$ column of $M$ and $N_j$ with the $j^{rm th}$ row of $j$. By the previous case, one of them has to be zero. So if $M$ has a nonzero column, then $N=0$; and if $N$ has a nonzero row, then $M=0$.






        share|cite|improve this answer














        No, it's not possible unless you accept that the range is ${0}$. Consider first the case $1,1$. We have
        $$
        begin{bmatrix} M_{11}\ M_{21}end{bmatrix}begin{bmatrix} N_{11}&N_{12}end{bmatrix}=begin{bmatrix} 0&a\a&0end{bmatrix}.
        $$
        Then $M_{11}N_{11}=0$. If $M_{11}ne0$, then $N_{11}=0$. But then $a=M_{21}N_{11}=0$. From the $1,2$ entry, $0=a=M_{11}N_{12}$, $N_{12}=0$ and $N=0$. When $N_{11}=0$, we obtain similarly that $M=0$ (just take transpose and apply the argument).



        Now in general, if $E_{kj}$ is the matrix with $1$ in the $k,j$ entry and zeroes elsewhere,
        $$
        ME_{kj}N=M_kN_j,
        $$
        where $M_k$ is the $k^{rm th}$ column of $M$ and $N_j$ with the $j^{rm th}$ row of $j$. By the previous case, one of them has to be zero. So if $M$ has a nonzero column, then $N=0$; and if $N$ has a nonzero row, then $M=0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 at 4:21

























        answered Nov 22 at 4:45









        Martin Argerami

        123k1176174




        123k1176174






























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