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Does the solvability of an equation $I=MV+NC$ for $V$ depend on $Ker(C)$?
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In a recently read paper I came across the statement that for matrices $MinRe^ntimesn_1$, $NinRe^ntimess_1$, $CinRe^s_1timesn$, the equation
beginequation
I=MV+NC tag1
endequation
has a solution for $VinRe^n_1timesn$ if the number of linear independent rows of $V$ must be at least equal to the dimension of $Ker(C)$. To show that the claim is true, lets take a vector $xinKer(C)$, then the above equation reduces to
beginequation
x=MVx=MbeginbmatrixV_1x\V_2x\vdots\V_n_1xendbmatrix tag2
endequation
where $V_i$ is the $i^th$ row of $V$. Since $Ker(C)neq0$ and thus $xneq0$ for which equation 2 holds if $MV=I$ or $rank(MV)leqn$. But I do not see the way forward. Any hint is greatly appreciated.
linear-algebra matrices linear-transformations matrix-equations
asked Jul 23 at 21:24
jbgujgu
1139
add a comment |Â
up vote
0
down vote
favorite
In a recently read paper I came across the statement that for matrices $MinRe^ntimesn_1$, $NinRe^ntimess_1$, $CinRe^s_1timesn$, the equation
beginequation
I=MV+NC tag1
endequation
has a solution for $VinRe^n_1timesn$ if the number of linear independent rows of $V$ must be at least equal to the dimension of $Ker(C)$. To show that the claim is true, lets take a vector $xinKer(C)$, then the above equation reduces to
beginequation
x=MVx=MbeginbmatrixV_1x\V_2x\vdots\V_n_1xendbmatrix tag2
endequation
where $V_i$ is the $i^th$ row of $V$. Since $Ker(C)neq0$ and thus $xneq0$ for which equation 2 holds if $MV=I$ or $rank(MV)leqn$. But I do not see the way forward. Any hint is greatly appreciated.
linear-algebra matrices linear-transformations matrix-equations
asked Jul 23 at 21:24
jbgujgu
1139
This isn't necessarily the case if $M = 0$ or $N = 0$. What information are we given? Do we know that $M$ and $N$ have full rank?
– Omnomnomnom
Jul 23 at 21:28
It is clear that the condition is necessary though, since $$ operatornamerank(MV + NC) leq operatornamerank(V) + operatornamerank(C) $$
– Omnomnomnom
Jul 23 at 21:29
Matrices $M$ and $N$ are non-zero but nothing is known for sure on the rank of $M$ and $N$. If these matrices are of full column rank, then your last comment holds.
– jbgujgu
Jul 23 at 21:33
My last comment holds whether or not $M$ and $N$ have full rank. However, if we're not guaranteed that $M,N$ have full rank, then no condition on $C,V$ will guarantee solvability.
– Omnomnomnom
Jul 23 at 21:35
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In a recently read paper I came across the statement that for matrices $MinRe^ntimesn_1$, $NinRe^ntimess_1$, $CinRe^s_1timesn$, the equation
beginequation
I=MV+NC tag1
endequation
has a solution for $VinRe^n_1timesn$ if the number of linear independent rows of $V$ must be at least equal to the dimension of $Ker(C)$. To show that the claim is true, lets take a vector $xinKer(C)$, then the above equation reduces to
beginequation
x=MVx=MbeginbmatrixV_1x\V_2x\vdots\V_n_1xendbmatrix tag2
endequation
where $V_i$ is the $i^th$ row of $V$. Since $Ker(C)neq0$ and thus $xneq0$ for which equation 2 holds if $MV=I$ or $rank(MV)leqn$. But I do not see the way forward. Any hint is greatly appreciated.
linear-algebra matrices linear-transformations matrix-equations
asked Jul 23 at 21:24
jbgujgu
1139
In a recently read paper I came across the statement that for matrices $MinRe^ntimesn_1$, $NinRe^ntimess_1$, $CinRe^s_1timesn$, the equation
beginequation
I=MV+NC tag1
endequation
has a solution for $VinRe^n_1timesn$ if the number of linear independent rows of $V$ must be at least equal to the dimension of $Ker(C)$. To show that the claim is true, lets take a vector $xinKer(C)$, then the above equation reduces to
beginequation
x=MVx=MbeginbmatrixV_1x\V_2x\vdots\V_n_1xendbmatrix tag2
endequation
where $V_i$ is the $i^th$ row of $V$. Since $Ker(C)neq0$ and thus $xneq0$ for which equation 2 holds if $MV=I$ or $rank(MV)leqn$. But I do not see the way forward. Any hint is greatly appreciated.
linear-algebra matrices linear-transformations matrix-equations
asked Jul 23 at 21:24
jbgujgu
1139
asked Jul 23 at 21:24
jbgujgu
1139
asked Jul 23 at 21:24
jbgujgu
1139
asked Jul 23 at 21:24
jbgujgu
1139
1139
This isn't necessarily the case if $M = 0$ or $N = 0$. What information are we given? Do we know that $M$ and $N$ have full rank?
– Omnomnomnom
Jul 23 at 21:28
It is clear that the condition is necessary though, since $$ operatornamerank(MV + NC) leq operatornamerank(V) + operatornamerank(C) $$
– Omnomnomnom
Jul 23 at 21:29
Matrices $M$ and $N$ are non-zero but nothing is known for sure on the rank of $M$ and $N$. If these matrices are of full column rank, then your last comment holds.
– jbgujgu
Jul 23 at 21:33
My last comment holds whether or not $M$ and $N$ have full rank. However, if we're not guaranteed that $M,N$ have full rank, then no condition on $C,V$ will guarantee solvability.
– Omnomnomnom
Jul 23 at 21:35
add a comment |Â
This isn't necessarily the case if $M = 0$ or $N = 0$. What information are we given? Do we know that $M$ and $N$ have full rank?
– Omnomnomnom
Jul 23 at 21:28
It is clear that the condition is necessary though, since $$ operatornamerank(MV + NC) leq operatornamerank(V) + operatornamerank(C) $$
– Omnomnomnom
Jul 23 at 21:29
Matrices $M$ and $N$ are non-zero but nothing is known for sure on the rank of $M$ and $N$. If these matrices are of full column rank, then your last comment holds.
– jbgujgu
Jul 23 at 21:33
My last comment holds whether or not $M$ and $N$ have full rank. However, if we're not guaranteed that $M,N$ have full rank, then no condition on $C,V$ will guarantee solvability.
– Omnomnomnom
Jul 23 at 21:35
This isn't necessarily the case if $M = 0$ or $N = 0$. What information are we given? Do we know that $M$ and $N$ have full rank?
– Omnomnomnom
Jul 23 at 21:28
This isn't necessarily the case if $M = 0$ or $N = 0$. What information are we given? Do we know that $M$ and $N$ have full rank?
– Omnomnomnom
Jul 23 at 21:28
It is clear that the condition is necessary though, since $$ operatornamerank(MV + NC) leq operatornamerank(V) + operatornamerank(C) $$
– Omnomnomnom
Jul 23 at 21:29
It is clear that the condition is necessary though, since $$ operatornamerank(MV + NC) leq operatornamerank(V) + operatornamerank(C) $$
– Omnomnomnom
Jul 23 at 21:29
Matrices $M$ and $N$ are non-zero but nothing is known for sure on the rank of $M$ and $N$. If these matrices are of full column rank, then your last comment holds.
– jbgujgu
Jul 23 at 21:33
Matrices $M$ and $N$ are non-zero but nothing is known for sure on the rank of $M$ and $N$. If these matrices are of full column rank, then your last comment holds.
– jbgujgu
Jul 23 at 21:33
My last comment holds whether or not $M$ and $N$ have full rank. However, if we're not guaranteed that $M,N$ have full rank, then no condition on $C,V$ will guarantee solvability.
– Omnomnomnom
Jul 23 at 21:35
My last comment holds whether or not $M$ and $N$ have full rank. However, if we're not guaranteed that $M,N$ have full rank, then no condition on $C,V$ will guarantee solvability.
– Omnomnomnom
Jul 23 at 21:35
add a comment |Â
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This isn't necessarily the case if $M = 0$ or $N = 0$. What information are we given? Do we know that $M$ and $N$ have full rank?
– Omnomnomnom
Jul 23 at 21:28
It is clear that the condition is necessary though, since $$ operatornamerank(MV + NC) leq operatornamerank(V) + operatornamerank(C) $$
– Omnomnomnom
Jul 23 at 21:29
Matrices $M$ and $N$ are non-zero but nothing is known for sure on the rank of $M$ and $N$. If these matrices are of full column rank, then your last comment holds.
– jbgujgu
Jul 23 at 21:33
My last comment holds whether or not $M$ and $N$ have full rank. However, if we're not guaranteed that $M,N$ have full rank, then no condition on $C,V$ will guarantee solvability.
– Omnomnomnom
Jul 23 at 21:35