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Is the set of matrices that has an eigenvector orthogonal to a fixed vector Zariski closed?

Topology property of the set of matrices that has an eigenvector orthogonal to a fixed vector.Join and Zariski closed setstopological properties of an algebraic set in the metric topologyProperties of a modified Zariski topologyThe Zariski closure of a constructible set is the same as the standard closure?Proving that principal open sets form a base for the Zariski topologyCompactness with the Zariski topologyDo irreducible sets have proper closed subsets in the Zariski topology?Is the set of real matrices diagonalizable in $mathcal M_n(C)$ dense in the set of all companion matrices?Topology property of the set of matrices that has an eigenvector orthogonal to a fixed vector.

2

$begingroup$

This is related to the question I asked.

Suppose $x in mathbb C^n$ is a fixed vector$.$ Define a set
$$beginalign*
mathcal E = A in M_n(mathbb C): exists text an eigen-pair (lambda, v) text of A, texti.e., Av = lambda v text such that v perp x .
endalign*$$

It is shown in the answer that $mathcal E$ is closed in the Euclidean topology. I want to know whether this set is closed in Zariski topology.

linear-algebra abstract-algebra algebraic-geometry zariski-topology

share|cite|improve this question

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

asked Mar 15 at 5:31

user43210user43210

678

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In general, I think the answer is negative. The condition of being “orthogonal complex vectors” cannot be expressed algebraically.
  $endgroup$
  – pyon
  Mar 15 at 5:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By “algebraically”, I mean “as the evaluation of a complex polynomial on the entries of a complex vector”.
  $endgroup$
  – pyon
  Mar 15 at 6:00
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @pyon: The role of the orthogonality here is trivial, though; it's just saying $A$ has an eigenvector in some particular subspace of $V$ (which happens to be the orthogonal complement of $x$).
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:30
  

  
  
  
  

add a comment | 

2

$begingroup$

This is related to the question I asked.

Suppose $x in mathbb C^n$ is a fixed vector$.$ Define a set
$$beginalign*
mathcal E = A in M_n(mathbb C): exists text an eigen-pair (lambda, v) text of A, texti.e., Av = lambda v text such that v perp x .
endalign*$$

It is shown in the answer that $mathcal E$ is closed in the Euclidean topology. I want to know whether this set is closed in Zariski topology.

linear-algebra abstract-algebra algebraic-geometry zariski-topology

share|cite|improve this question

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

asked Mar 15 at 5:31

user43210user43210

678

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In general, I think the answer is negative. The condition of being “orthogonal complex vectors” cannot be expressed algebraically.
  $endgroup$
  – pyon
  Mar 15 at 5:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By “algebraically”, I mean “as the evaluation of a complex polynomial on the entries of a complex vector”.
  $endgroup$
  – pyon
  Mar 15 at 6:00
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @pyon: The role of the orthogonality here is trivial, though; it's just saying $A$ has an eigenvector in some particular subspace of $V$ (which happens to be the orthogonal complement of $x$).
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:30
  

  
  
  
  

add a comment | 

$begingroup$

This is related to the question I asked.

Suppose $x in mathbb C^n$ is a fixed vector$.$ Define a set
$$beginalign*
mathcal E = A in M_n(mathbb C): exists text an eigen-pair (lambda, v) text of A, texti.e., Av = lambda v text such that v perp x .
endalign*$$

It is shown in the answer that $mathcal E$ is closed in the Euclidean topology. I want to know whether this set is closed in Zariski topology.

linear-algebra abstract-algebra algebraic-geometry zariski-topology

share|cite|improve this question

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

asked Mar 15 at 5:31

user43210user43210

678

$endgroup$

share|cite|improve this question

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

asked Mar 15 at 5:31

user43210user43210

678

share|cite|improve this question

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

asked Mar 15 at 5:31

user43210user43210

678

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

edited Mar 15 at 6:34

Eric Wofsey

190k14216348

asked Mar 15 at 5:31

user43210user43210

678

asked Mar 15 at 5:31

user43210user43210

678

1 Answer
1

active

oldest

votes

3

$begingroup$

Yes. We may assume $x=(0,0,dots,1)$. Then $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$. This happens for a particular value of $lambda$ iff all of the $(n-1)times(n-1)$ minors of the first $n-1$ columns of $lambda I-A$ vanish. Each of these minors is a polynomial $p_k(A,lambda)$ in the entries of $A$ and $lambda$, and we are looking for the set of $A$ such that these polynomials have a common root in $lambda$.

Let $X_ksubseteq M_n(mathbbC)times mathbbC$ be the vanishing set of the $k$th minor $p_k(A,lambda)$ (obviously a Zariski closed set). Then $mathcalE$ is the projection of $bigcap X_k$ onto $M_n(mathbbC)$. Let $Y_k$ be the Zariski closure of $X_k$ in $M_n(mathbbC)timesmathbbP^1$, or equivalently the vanishing set of the polynomial $q_k(A,s,t)=t^dp_k(A,s/t)$ where $d$ is the degree of $p_k$ with respect to $lambda$ (here $[s:t]$ are the homogeneous coordinates on $mathbbP^1$). Note that the minor $p_n$ formed by omitting the bottom row is monic in $lambda$, and so $q_n$ does not vanish at $t=0$ and thus $X_n=Y_n$. It follows that $bigcap Y_k=bigcap X_k$. But the projection of $bigcap Y_k$ to $M_n(mathbbC)$ is Zariski-closed, since the projection $M_n(mathbbC)timesmathbbP^1to M_n(mathbbC)$ is a closed map in the Zariski topology. Since $bigcap Y_k=bigcap X_k$, this means $mathcalE$ is Zariski-closed.

share|cite|improve this answer

edited Mar 15 at 6:37

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. Could you explain a little bit why $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$?
  $endgroup$
  – user43210
  Mar 15 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  An element of the kernel of the matrix formed by those first $n-1$ columns is the same thing as an eigenvector of $A$ with eigenvalue $lambda$ whose last entry is $0$.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. Thanks. So the complement of $mathcal E$ is Zariski open. I confuse myself when thinking about the complement. Is this correct that $mathcal E^c = A in M_n(mathbb C): forall text eigen-pair (lambda, v), , v notperp x$?
  $endgroup$
  – user43210
  Mar 15 at 6:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's correct.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is the projection $M_n(mathbb C) times mathbb P^1 to M_n(mathbb C)$ closed? Does the conclusion change if replace the field to be $mathbb R$ where $(lambda, v)$ is still considered in $mathbb C$? Thanks. Forgive me if the question is too stupid (I only start on Zariski topology).
  $endgroup$
  – user43210
  Mar 15 at 7:21
  

  
  
  
  

 | 
show 5 more comments

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3

$begingroup$

Yes. We may assume $x=(0,0,dots,1)$. Then $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$. This happens for a particular value of $lambda$ iff all of the $(n-1)times(n-1)$ minors of the first $n-1$ columns of $lambda I-A$ vanish. Each of these minors is a polynomial $p_k(A,lambda)$ in the entries of $A$ and $lambda$, and we are looking for the set of $A$ such that these polynomials have a common root in $lambda$.

Let $X_ksubseteq M_n(mathbbC)times mathbbC$ be the vanishing set of the $k$th minor $p_k(A,lambda)$ (obviously a Zariski closed set). Then $mathcalE$ is the projection of $bigcap X_k$ onto $M_n(mathbbC)$. Let $Y_k$ be the Zariski closure of $X_k$ in $M_n(mathbbC)timesmathbbP^1$, or equivalently the vanishing set of the polynomial $q_k(A,s,t)=t^dp_k(A,s/t)$ where $d$ is the degree of $p_k$ with respect to $lambda$ (here $[s:t]$ are the homogeneous coordinates on $mathbbP^1$). Note that the minor $p_n$ formed by omitting the bottom row is monic in $lambda$, and so $q_n$ does not vanish at $t=0$ and thus $X_n=Y_n$. It follows that $bigcap Y_k=bigcap X_k$. But the projection of $bigcap Y_k$ to $M_n(mathbbC)$ is Zariski-closed, since the projection $M_n(mathbbC)timesmathbbP^1to M_n(mathbbC)$ is a closed map in the Zariski topology. Since $bigcap Y_k=bigcap X_k$, this means $mathcalE$ is Zariski-closed.

share|cite|improve this answer

edited Mar 15 at 6:37

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. Could you explain a little bit why $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$?
  $endgroup$
  – user43210
  Mar 15 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  An element of the kernel of the matrix formed by those first $n-1$ columns is the same thing as an eigenvector of $A$ with eigenvalue $lambda$ whose last entry is $0$.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. Thanks. So the complement of $mathcal E$ is Zariski open. I confuse myself when thinking about the complement. Is this correct that $mathcal E^c = A in M_n(mathbb C): forall text eigen-pair (lambda, v), , v notperp x$?
  $endgroup$
  – user43210
  Mar 15 at 6:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's correct.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is the projection $M_n(mathbb C) times mathbb P^1 to M_n(mathbb C)$ closed? Does the conclusion change if replace the field to be $mathbb R$ where $(lambda, v)$ is still considered in $mathbb C$? Thanks. Forgive me if the question is too stupid (I only start on Zariski topology).
  $endgroup$
  – user43210
  Mar 15 at 7:21
  

  
  
  
  

 | 
show 5 more comments

3

$begingroup$

Yes. We may assume $x=(0,0,dots,1)$. Then $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$. This happens for a particular value of $lambda$ iff all of the $(n-1)times(n-1)$ minors of the first $n-1$ columns of $lambda I-A$ vanish. Each of these minors is a polynomial $p_k(A,lambda)$ in the entries of $A$ and $lambda$, and we are looking for the set of $A$ such that these polynomials have a common root in $lambda$.

Let $X_ksubseteq M_n(mathbbC)times mathbbC$ be the vanishing set of the $k$th minor $p_k(A,lambda)$ (obviously a Zariski closed set). Then $mathcalE$ is the projection of $bigcap X_k$ onto $M_n(mathbbC)$. Let $Y_k$ be the Zariski closure of $X_k$ in $M_n(mathbbC)timesmathbbP^1$, or equivalently the vanishing set of the polynomial $q_k(A,s,t)=t^dp_k(A,s/t)$ where $d$ is the degree of $p_k$ with respect to $lambda$ (here $[s:t]$ are the homogeneous coordinates on $mathbbP^1$). Note that the minor $p_n$ formed by omitting the bottom row is monic in $lambda$, and so $q_n$ does not vanish at $t=0$ and thus $X_n=Y_n$. It follows that $bigcap Y_k=bigcap X_k$. But the projection of $bigcap Y_k$ to $M_n(mathbbC)$ is Zariski-closed, since the projection $M_n(mathbbC)timesmathbbP^1to M_n(mathbbC)$ is a closed map in the Zariski topology. Since $bigcap Y_k=bigcap X_k$, this means $mathcalE$ is Zariski-closed.

share|cite|improve this answer

edited Mar 15 at 6:37

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. Could you explain a little bit why $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$?
  $endgroup$
  – user43210
  Mar 15 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  An element of the kernel of the matrix formed by those first $n-1$ columns is the same thing as an eigenvector of $A$ with eigenvalue $lambda$ whose last entry is $0$.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. Thanks. So the complement of $mathcal E$ is Zariski open. I confuse myself when thinking about the complement. Is this correct that $mathcal E^c = A in M_n(mathbb C): forall text eigen-pair (lambda, v), , v notperp x$?
  $endgroup$
  – user43210
  Mar 15 at 6:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's correct.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is the projection $M_n(mathbb C) times mathbb P^1 to M_n(mathbb C)$ closed? Does the conclusion change if replace the field to be $mathbb R$ where $(lambda, v)$ is still considered in $mathbb C$? Thanks. Forgive me if the question is too stupid (I only start on Zariski topology).
  $endgroup$
  – user43210
  Mar 15 at 7:21
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Yes. We may assume $x=(0,0,dots,1)$. Then $mathcalE$ is the set of $A$ such that there exists $lambdainmathbbC$ such that the matrix formed by the first $n-1$ columns of $lambda I-A$ has rank less than $n-1$. This happens for a particular value of $lambda$ iff all of the $(n-1)times(n-1)$ minors of the first $n-1$ columns of $lambda I-A$ vanish. Each of these minors is a polynomial $p_k(A,lambda)$ in the entries of $A$ and $lambda$, and we are looking for the set of $A$ such that these polynomials have a common root in $lambda$.

Let $X_ksubseteq M_n(mathbbC)times mathbbC$ be the vanishing set of the $k$th minor $p_k(A,lambda)$ (obviously a Zariski closed set). Then $mathcalE$ is the projection of $bigcap X_k$ onto $M_n(mathbbC)$. Let $Y_k$ be the Zariski closure of $X_k$ in $M_n(mathbbC)timesmathbbP^1$, or equivalently the vanishing set of the polynomial $q_k(A,s,t)=t^dp_k(A,s/t)$ where $d$ is the degree of $p_k$ with respect to $lambda$ (here $[s:t]$ are the homogeneous coordinates on $mathbbP^1$). Note that the minor $p_n$ formed by omitting the bottom row is monic in $lambda$, and so $q_n$ does not vanish at $t=0$ and thus $X_n=Y_n$. It follows that $bigcap Y_k=bigcap X_k$. But the projection of $bigcap Y_k$ to $M_n(mathbbC)$ is Zariski-closed, since the projection $M_n(mathbbC)timesmathbbP^1to M_n(mathbbC)$ is a closed map in the Zariski topology. Since $bigcap Y_k=bigcap X_k$, this means $mathcalE$ is Zariski-closed.

share|cite|improve this answer

edited Mar 15 at 6:37

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348

$endgroup$

share|cite|improve this answer

edited Mar 15 at 6:37

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348

answered Mar 15 at 6:29

Eric WofseyEric Wofsey

190k14216348