Is it possible to find matrices $M, N$ such that $MXN$ has specified structure for all $X$?
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
asked Nov 22 at 3:26
user1101010
7551630
add a comment |
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
asked Nov 22 at 3:26
user1101010
7551630
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
add a comment |
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
asked Nov 22 at 3:26
user1101010
7551630
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
asked Nov 22 at 3:26
user1101010
7551630
asked Nov 22 at 3:26
user1101010
7551630
edited Nov 22 at 3:41
asked Nov 22 at 3:26
user1101010
7551630
asked Nov 22 at 3:26
user1101010
7551630
asked Nov 22 at 3:26
user1101010
7551630
7551630
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
add a comment |
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
add a comment |
1 Answer
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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 $M=0$.
answered Nov 22 at 4:45
Martin Argerami
123k1176174
add a comment |
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1 Answer
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1 Answer
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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 $M=0$.
answered Nov 22 at 4:45
Martin Argerami
123k1176174
add a comment |
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 $M=0$.
answered Nov 22 at 4:45
Martin Argerami
123k1176174
add a comment |
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 $M=0$.
answered Nov 22 at 4:45
Martin Argerami
123k1176174
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 $M=0$.
answered Nov 22 at 4:45
Martin Argerami
123k1176174
edited Nov 26 at 4:21
answered Nov 22 at 4:45
Martin Argerami
123k1176174
answered Nov 22 at 4:45
Martin Argerami
123k1176174
answered Nov 22 at 4:45
Martin Argerami
123k1176174
123k1176174
add a comment |
add a comment |
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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