A sort of weak nullstellensatz for matrices
$begingroup$
Be $S$ a finite set of square matrices $A_i$ of order $n$ with entries in $mathbb C$, such that the null vector is the only element in the intersection of their kernels: $$cap_i ker(A_i)={0}$$
Does the left ideal generated by $S$ coincide with the unit ideal of $M_n(mathbb C)$?
linear-algebra matrices
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
$endgroup$
|
show 1 more comment
$begingroup$
Be $S$ a finite set of square matrices $A_i$ of order $n$ with entries in $mathbb C$, such that the null vector is the only element in the intersection of their kernels: $$cap_i ker(A_i)={0}$$
Does the left ideal generated by $S$ coincide with the unit ideal of $M_n(mathbb C)$?
linear-algebra matrices
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
$endgroup$
$begingroup$
It's not true in general: take $S$ to contain only one element, $A$, which is a matrix of full rank that is not a scalar multiple of the identity. Then $ker(A) = {0}$, but $A$ does not generate the unit ideal.
$endgroup$
– user3482749
Dec 28 '18 at 15:54
$begingroup$
If a matrix $A$ is of full rank then exist $B$ and $C$ such that $$BAC=1$$ So $A$ does generates the unit ideal. Where am i wrong?
$endgroup$
– Abdenier
Dec 28 '18 at 16:35
$begingroup$
Sorry, I got slightly confused as to in what sense it was generating it.
$endgroup$
– user3482749
Dec 28 '18 at 19:04
$begingroup$
I think the answer is that, as long as we deal with two-sided ideals intersection of kernels doesn't matter and the condition for generating the unit ideal is that the sum of ranks is at least $n$. My wonder now is about one-sided ideals..
$endgroup$
– Abdenier
Dec 28 '18 at 19:39
$begingroup$
You need some kind of "general position" type hypothesis in there (else take $n$ scalar multiples of a matrix of rank 1).
$endgroup$
– user3482749
Dec 28 '18 at 19:54
|
show 1 more comment
$begingroup$
Be $S$ a finite set of square matrices $A_i$ of order $n$ with entries in $mathbb C$, such that the null vector is the only element in the intersection of their kernels: $$cap_i ker(A_i)={0}$$
Does the left ideal generated by $S$ coincide with the unit ideal of $M_n(mathbb C)$?
linear-algebra matrices
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
$endgroup$
Be $S$ a finite set of square matrices $A_i$ of order $n$ with entries in $mathbb C$, such that the null vector is the only element in the intersection of their kernels: $$cap_i ker(A_i)={0}$$
Does the left ideal generated by $S$ coincide with the unit ideal of $M_n(mathbb C)$?
linear-algebra matrices
linear-algebra matrices
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
edited Dec 28 '18 at 22:37
Abdenier
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
asked Dec 28 '18 at 15:47
AbdenierAbdenier
133
133
$begingroup$
It's not true in general: take $S$ to contain only one element, $A$, which is a matrix of full rank that is not a scalar multiple of the identity. Then $ker(A) = {0}$, but $A$ does not generate the unit ideal.
$endgroup$
– user3482749
Dec 28 '18 at 15:54
$begingroup$
If a matrix $A$ is of full rank then exist $B$ and $C$ such that $$BAC=1$$ So $A$ does generates the unit ideal. Where am i wrong?
$endgroup$
– Abdenier
Dec 28 '18 at 16:35
$begingroup$
Sorry, I got slightly confused as to in what sense it was generating it.
$endgroup$
– user3482749
Dec 28 '18 at 19:04
$begingroup$
I think the answer is that, as long as we deal with two-sided ideals intersection of kernels doesn't matter and the condition for generating the unit ideal is that the sum of ranks is at least $n$. My wonder now is about one-sided ideals..
$endgroup$
– Abdenier
Dec 28 '18 at 19:39
$begingroup$
You need some kind of "general position" type hypothesis in there (else take $n$ scalar multiples of a matrix of rank 1).
$endgroup$
– user3482749
Dec 28 '18 at 19:54
|
show 1 more comment
$begingroup$
It's not true in general: take $S$ to contain only one element, $A$, which is a matrix of full rank that is not a scalar multiple of the identity. Then $ker(A) = {0}$, but $A$ does not generate the unit ideal.
$endgroup$
– user3482749
Dec 28 '18 at 15:54
$begingroup$
If a matrix $A$ is of full rank then exist $B$ and $C$ such that $$BAC=1$$ So $A$ does generates the unit ideal. Where am i wrong?
$endgroup$
– Abdenier
Dec 28 '18 at 16:35
$begingroup$
Sorry, I got slightly confused as to in what sense it was generating it.
$endgroup$
– user3482749
Dec 28 '18 at 19:04
$begingroup$
I think the answer is that, as long as we deal with two-sided ideals intersection of kernels doesn't matter and the condition for generating the unit ideal is that the sum of ranks is at least $n$. My wonder now is about one-sided ideals..
$endgroup$
– Abdenier
Dec 28 '18 at 19:39
$begingroup$
You need some kind of "general position" type hypothesis in there (else take $n$ scalar multiples of a matrix of rank 1).
$endgroup$
– user3482749
Dec 28 '18 at 19:54
$begingroup$
It's not true in general: take $S$ to contain only one element, $A$, which is a matrix of full rank that is not a scalar multiple of the identity. Then $ker(A) = {0}$, but $A$ does not generate the unit ideal.
$endgroup$
– user3482749
Dec 28 '18 at 15:54
$begingroup$
It's not true in general: take $S$ to contain only one element, $A$, which is a matrix of full rank that is not a scalar multiple of the identity. Then $ker(A) = {0}$, but $A$ does not generate the unit ideal.
$endgroup$
– user3482749
Dec 28 '18 at 15:54
$begingroup$
If a matrix $A$ is of full rank then exist $B$ and $C$ such that $$BAC=1$$ So $A$ does generates the unit ideal. Where am i wrong?
$endgroup$
– Abdenier
Dec 28 '18 at 16:35
$begingroup$
If a matrix $A$ is of full rank then exist $B$ and $C$ such that $$BAC=1$$ So $A$ does generates the unit ideal. Where am i wrong?
$endgroup$
– Abdenier
Dec 28 '18 at 16:35
$begingroup$
Sorry, I got slightly confused as to in what sense it was generating it.
$endgroup$
– user3482749
Dec 28 '18 at 19:04
$begingroup$
Sorry, I got slightly confused as to in what sense it was generating it.
$endgroup$
– user3482749
Dec 28 '18 at 19:04
$begingroup$
I think the answer is that, as long as we deal with two-sided ideals intersection of kernels doesn't matter and the condition for generating the unit ideal is that the sum of ranks is at least $n$. My wonder now is about one-sided ideals..
$endgroup$
– Abdenier
Dec 28 '18 at 19:39
$begingroup$
I think the answer is that, as long as we deal with two-sided ideals intersection of kernels doesn't matter and the condition for generating the unit ideal is that the sum of ranks is at least $n$. My wonder now is about one-sided ideals..
$endgroup$
– Abdenier
Dec 28 '18 at 19:39
$begingroup$
You need some kind of "general position" type hypothesis in there (else take $n$ scalar multiples of a matrix of rank 1).
$endgroup$
– user3482749
Dec 28 '18 at 19:54
$begingroup$
You need some kind of "general position" type hypothesis in there (else take $n$ scalar multiples of a matrix of rank 1).
$endgroup$
– user3482749
Dec 28 '18 at 19:54
|
show 1 more comment
1 Answer
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$begingroup$
Yes, this is true, and works over any field $k$. Let $Isubseteq M_n(k)$ be a left ideal. Since $M_n(k)$ is a semisimple ring, there is another left ideal $Jsubseteq M_n(k)$ such that $Ioplus J=M_n(k)$. Let $iin I$ and $jin J$ be such that $i+j=1$. Then for any $i'in I$, $i'j=i'(1-i)in Icap J=0$, so $i'=i'i$. In particular, $ker(i)subseteq ker(i')$ for all $i'in I$.
Now in your case that $I$ is generated by matrices whose kernels have trivial intersection, this means that $ker(i)$ must be trivial. Thus $i$ is invertible, so $I=M_n(k)$ since $iin I$.
(More generally, we can conclude that every left ideal in $M_n(k)$ the set of matrices that vanish on some subspace of $k^n$, namely the kernel of $i$.)
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
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$begingroup$
Yes, this is true, and works over any field $k$. Let $Isubseteq M_n(k)$ be a left ideal. Since $M_n(k)$ is a semisimple ring, there is another left ideal $Jsubseteq M_n(k)$ such that $Ioplus J=M_n(k)$. Let $iin I$ and $jin J$ be such that $i+j=1$. Then for any $i'in I$, $i'j=i'(1-i)in Icap J=0$, so $i'=i'i$. In particular, $ker(i)subseteq ker(i')$ for all $i'in I$.
Now in your case that $I$ is generated by matrices whose kernels have trivial intersection, this means that $ker(i)$ must be trivial. Thus $i$ is invertible, so $I=M_n(k)$ since $iin I$.
(More generally, we can conclude that every left ideal in $M_n(k)$ the set of matrices that vanish on some subspace of $k^n$, namely the kernel of $i$.)
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
Yes, this is true, and works over any field $k$. Let $Isubseteq M_n(k)$ be a left ideal. Since $M_n(k)$ is a semisimple ring, there is another left ideal $Jsubseteq M_n(k)$ such that $Ioplus J=M_n(k)$. Let $iin I$ and $jin J$ be such that $i+j=1$. Then for any $i'in I$, $i'j=i'(1-i)in Icap J=0$, so $i'=i'i$. In particular, $ker(i)subseteq ker(i')$ for all $i'in I$.
Now in your case that $I$ is generated by matrices whose kernels have trivial intersection, this means that $ker(i)$ must be trivial. Thus $i$ is invertible, so $I=M_n(k)$ since $iin I$.
(More generally, we can conclude that every left ideal in $M_n(k)$ the set of matrices that vanish on some subspace of $k^n$, namely the kernel of $i$.)
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
Yes, this is true, and works over any field $k$. Let $Isubseteq M_n(k)$ be a left ideal. Since $M_n(k)$ is a semisimple ring, there is another left ideal $Jsubseteq M_n(k)$ such that $Ioplus J=M_n(k)$. Let $iin I$ and $jin J$ be such that $i+j=1$. Then for any $i'in I$, $i'j=i'(1-i)in Icap J=0$, so $i'=i'i$. In particular, $ker(i)subseteq ker(i')$ for all $i'in I$.
Now in your case that $I$ is generated by matrices whose kernels have trivial intersection, this means that $ker(i)$ must be trivial. Thus $i$ is invertible, so $I=M_n(k)$ since $iin I$.
(More generally, we can conclude that every left ideal in $M_n(k)$ the set of matrices that vanish on some subspace of $k^n$, namely the kernel of $i$.)
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
$endgroup$
Yes, this is true, and works over any field $k$. Let $Isubseteq M_n(k)$ be a left ideal. Since $M_n(k)$ is a semisimple ring, there is another left ideal $Jsubseteq M_n(k)$ such that $Ioplus J=M_n(k)$. Let $iin I$ and $jin J$ be such that $i+j=1$. Then for any $i'in I$, $i'j=i'(1-i)in Icap J=0$, so $i'=i'i$. In particular, $ker(i)subseteq ker(i')$ for all $i'in I$.
Now in your case that $I$ is generated by matrices whose kernels have trivial intersection, this means that $ker(i)$ must be trivial. Thus $i$ is invertible, so $I=M_n(k)$ since $iin I$.
(More generally, we can conclude that every left ideal in $M_n(k)$ the set of matrices that vanish on some subspace of $k^n$, namely the kernel of $i$.)
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
answered Dec 28 '18 at 21:33
Eric WofseyEric Wofsey
193k14221352
193k14221352
add a comment |
add a comment |
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$begingroup$
It's not true in general: take $S$ to contain only one element, $A$, which is a matrix of full rank that is not a scalar multiple of the identity. Then $ker(A) = {0}$, but $A$ does not generate the unit ideal.
$endgroup$
– user3482749
Dec 28 '18 at 15:54
$begingroup$
If a matrix $A$ is of full rank then exist $B$ and $C$ such that $$BAC=1$$ So $A$ does generates the unit ideal. Where am i wrong?
$endgroup$
– Abdenier
Dec 28 '18 at 16:35
$begingroup$
Sorry, I got slightly confused as to in what sense it was generating it.
$endgroup$
– user3482749
Dec 28 '18 at 19:04
$begingroup$
I think the answer is that, as long as we deal with two-sided ideals intersection of kernels doesn't matter and the condition for generating the unit ideal is that the sum of ranks is at least $n$. My wonder now is about one-sided ideals..
$endgroup$
– Abdenier
Dec 28 '18 at 19:39
$begingroup$
You need some kind of "general position" type hypothesis in there (else take $n$ scalar multiples of a matrix of rank 1).
$endgroup$
– user3482749
Dec 28 '18 at 19:54