Codimension of a subset of $Hom(E,F)$
$begingroup$
Let $E$ and $F$ be vector spaces over the field of complex numbers. Consider
$$W={fin Hom(E,F),:, textrm{dim }ker(f)geq c},.$$
The claim is that $W$ is a closed subvariety of $Hom(E,F)$ of codimension $max{0,c(m+c)}$ where $m=text{dim } F-text{dim }E$. How do we prove this?
I have first been trying to prove the case when $E=F$. In this case I need to prove that $W$ is of codimension $c^2$ in $Hom(E,E)$. I need to cut down by $c^2$ equations. Are they equations of certain minors?
algebraic-geometry affine-geometry
asked Dec 19 '18 at 4:31
user52991user52991
341310
$endgroup$
add a comment |
$begingroup$
Let $E$ and $F$ be vector spaces over the field of complex numbers. Consider
$$W={fin Hom(E,F),:, textrm{dim }ker(f)geq c},.$$
The claim is that $W$ is a closed subvariety of $Hom(E,F)$ of codimension $max{0,c(m+c)}$ where $m=text{dim } F-text{dim }E$. How do we prove this?
I have first been trying to prove the case when $E=F$. In this case I need to prove that $W$ is of codimension $c^2$ in $Hom(E,E)$. I need to cut down by $c^2$ equations. Are they equations of certain minors?
algebraic-geometry affine-geometry
asked Dec 19 '18 at 4:31
user52991user52991
341310
$endgroup$
add a comment |
$begingroup$
Let $E$ and $F$ be vector spaces over the field of complex numbers. Consider
$$W={fin Hom(E,F),:, textrm{dim }ker(f)geq c},.$$
The claim is that $W$ is a closed subvariety of $Hom(E,F)$ of codimension $max{0,c(m+c)}$ where $m=text{dim } F-text{dim }E$. How do we prove this?
I have first been trying to prove the case when $E=F$. In this case I need to prove that $W$ is of codimension $c^2$ in $Hom(E,E)$. I need to cut down by $c^2$ equations. Are they equations of certain minors?
algebraic-geometry affine-geometry
asked Dec 19 '18 at 4:31
user52991user52991
341310
$endgroup$
Let $E$ and $F$ be vector spaces over the field of complex numbers. Consider
$$W={fin Hom(E,F),:, textrm{dim }ker(f)geq c},.$$
The claim is that $W$ is a closed subvariety of $Hom(E,F)$ of codimension $max{0,c(m+c)}$ where $m=text{dim } F-text{dim }E$. How do we prove this?
I have first been trying to prove the case when $E=F$. In this case I need to prove that $W$ is of codimension $c^2$ in $Hom(E,E)$. I need to cut down by $c^2$ equations. Are they equations of certain minors?
algebraic-geometry affine-geometry
algebraic-geometry affine-geometry
asked Dec 19 '18 at 4:31
user52991user52991
341310
asked Dec 19 '18 at 4:31
user52991user52991
341310
asked Dec 19 '18 at 4:31
user52991user52991
341310
asked Dec 19 '18 at 4:31
user52991user52991
341310
asked Dec 19 '18 at 4:31
user52991user52991
341310
341310
add a comment |
add a comment |
1 Answer
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Yes, essentially. If $dim ker fge c$, let $r=dim E-c$, so $dimker f ge c$ is equivalent to $operatorname{rank} f le r$. Then any $r+1times r+1$ minor of $f$ must be $0$, and conversely if every $r+1times r+1$ minor of $f$ is $0$, then the rank of $f$ is at most $r$, so the kernel is at least $c$ dimensional.
How many minors are there? Well if $n=dim E$, then there are $(n-r)(n+m-r)=c(m+c)$ minors, giving the desired codimension. I assume that you can handle the edge cases ofc.
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
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$begingroup$
Yes, essentially. If $dim ker fge c$, let $r=dim E-c$, so $dimker f ge c$ is equivalent to $operatorname{rank} f le r$. Then any $r+1times r+1$ minor of $f$ must be $0$, and conversely if every $r+1times r+1$ minor of $f$ is $0$, then the rank of $f$ is at most $r$, so the kernel is at least $c$ dimensional.
How many minors are there? Well if $n=dim E$, then there are $(n-r)(n+m-r)=c(m+c)$ minors, giving the desired codimension. I assume that you can handle the edge cases ofc.
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
$endgroup$
add a comment |
$begingroup$
Yes, essentially. If $dim ker fge c$, let $r=dim E-c$, so $dimker f ge c$ is equivalent to $operatorname{rank} f le r$. Then any $r+1times r+1$ minor of $f$ must be $0$, and conversely if every $r+1times r+1$ minor of $f$ is $0$, then the rank of $f$ is at most $r$, so the kernel is at least $c$ dimensional.
How many minors are there? Well if $n=dim E$, then there are $(n-r)(n+m-r)=c(m+c)$ minors, giving the desired codimension. I assume that you can handle the edge cases ofc.
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
$endgroup$
add a comment |
$begingroup$
Yes, essentially. If $dim ker fge c$, let $r=dim E-c$, so $dimker f ge c$ is equivalent to $operatorname{rank} f le r$. Then any $r+1times r+1$ minor of $f$ must be $0$, and conversely if every $r+1times r+1$ minor of $f$ is $0$, then the rank of $f$ is at most $r$, so the kernel is at least $c$ dimensional.
How many minors are there? Well if $n=dim E$, then there are $(n-r)(n+m-r)=c(m+c)$ minors, giving the desired codimension. I assume that you can handle the edge cases ofc.
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
$endgroup$
Yes, essentially. If $dim ker fge c$, let $r=dim E-c$, so $dimker f ge c$ is equivalent to $operatorname{rank} f le r$. Then any $r+1times r+1$ minor of $f$ must be $0$, and conversely if every $r+1times r+1$ minor of $f$ is $0$, then the rank of $f$ is at most $r$, so the kernel is at least $c$ dimensional.
How many minors are there? Well if $n=dim E$, then there are $(n-r)(n+m-r)=c(m+c)$ minors, giving the desired codimension. I assume that you can handle the edge cases ofc.
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
answered Dec 19 '18 at 15:33
jgonjgon
14.7k22042
14.7k22042
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