connectedness of conjugacy classes of a fixed matrix $A$ but with the first column invariant

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Let $A in M_n(mathbb C)$ be a fixed matrix. The set $S^-1 A S: S in GL_n(mathbb C)$ is a continuous image of $GL_n(mathbb C)$, so I think it should be a connected set. Let $A = (a_1, dots, a_n)$ where $a_j in mathbb C^n$ denotes the columns of $A$. Let $F = S^-1 A S: S in GL_n(mathbb C) text and (S^-1AS)_cdot,1 = a_1$. Is the set $F$ still connected?

linear-algebra abstract-algebra general-topology connectedness

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asked Jul 18 at 20:19

user9527

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up vote
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Let $A in M_n(mathbb C)$ be a fixed matrix. The set $S^-1 A S: S in GL_n(mathbb C)$ is a continuous image of $GL_n(mathbb C)$, so I think it should be a connected set. Let $A = (a_1, dots, a_n)$ where $a_j in mathbb C^n$ denotes the columns of $A$. Let $F = S^-1 A S: S in GL_n(mathbb C) text and (S^-1AS)_cdot,1 = a_1$. Is the set $F$ still connected?

linear-algebra abstract-algebra general-topology connectedness

share|cite|improve this question

asked Jul 18 at 20:19

user9527

925525

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $A in M_n(mathbb C)$ be a fixed matrix. The set $S^-1 A S: S in GL_n(mathbb C)$ is a continuous image of $GL_n(mathbb C)$, so I think it should be a connected set. Let $A = (a_1, dots, a_n)$ where $a_j in mathbb C^n$ denotes the columns of $A$. Let $F = S^-1 A S: S in GL_n(mathbb C) text and (S^-1AS)_cdot,1 = a_1$. Is the set $F$ still connected?

linear-algebra abstract-algebra general-topology connectedness

share|cite|improve this question

asked Jul 18 at 20:19

user9527

925525

Let $A in M_n(mathbb C)$ be a fixed matrix. The set $S^-1 A S: S in GL_n(mathbb C)$ is a continuous image of $GL_n(mathbb C)$, so I think it should be a connected set. Let $A = (a_1, dots, a_n)$ where $a_j in mathbb C^n$ denotes the columns of $A$. Let $F = S^-1 A S: S in GL_n(mathbb C) text and (S^-1AS)_cdot,1 = a_1$. Is the set $F$ still connected?

linear-algebra abstract-algebra general-topology connectedness

share|cite|improve this question

asked Jul 18 at 20:19

user9527

925525

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 20:19

user9527

925525

asked Jul 18 at 20:19

user9527

925525

asked Jul 18 at 20:19

user9527

925525

925525

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Your condition $(S^-1AS)_cdot,1 = a_1$ is equivalent to $(SA-AS)e_1 = 0$.
Let $S$ be a matrix in your set and let $gamma : [0,1]tomathbb C$ be a path such that $gamma(0) = 0$, $gamma(1) = 1$, and $det(S + gamma(t)(I-S))neq 0$. It is now easy to see that $S(t) = S + gamma(t)(I-S)$ is in your set for each $tin [0,1]$ with $S(0) = S$ and $S(1) = I$. So, we have found a path within your set from $S$ to $I$. This shows that it is connected.

share|cite|improve this answer

answered Jul 18 at 21:21

amsmath

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
  – user9527
  Jul 18 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
  – amsmath
  Jul 18 at 21:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your nice answer.
  – user9527
  Jul 18 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
  – user9527
  Jul 19 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $GL(n,mathbb R)$ is not connected.
  – amsmath
  Jul 19 at 20:20
  

  
  
  
  

 | 
show 2 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Your condition $(S^-1AS)_cdot,1 = a_1$ is equivalent to $(SA-AS)e_1 = 0$.
Let $S$ be a matrix in your set and let $gamma : [0,1]tomathbb C$ be a path such that $gamma(0) = 0$, $gamma(1) = 1$, and $det(S + gamma(t)(I-S))neq 0$. It is now easy to see that $S(t) = S + gamma(t)(I-S)$ is in your set for each $tin [0,1]$ with $S(0) = S$ and $S(1) = I$. So, we have found a path within your set from $S$ to $I$. This shows that it is connected.

share|cite|improve this answer

answered Jul 18 at 21:21

amsmath

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
  – user9527
  Jul 18 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
  – amsmath
  Jul 18 at 21:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your nice answer.
  – user9527
  Jul 18 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
  – user9527
  Jul 19 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $GL(n,mathbb R)$ is not connected.
  – amsmath
  Jul 19 at 20:20
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote



accepted

Your condition $(S^-1AS)_cdot,1 = a_1$ is equivalent to $(SA-AS)e_1 = 0$.
Let $S$ be a matrix in your set and let $gamma : [0,1]tomathbb C$ be a path such that $gamma(0) = 0$, $gamma(1) = 1$, and $det(S + gamma(t)(I-S))neq 0$. It is now easy to see that $S(t) = S + gamma(t)(I-S)$ is in your set for each $tin [0,1]$ with $S(0) = S$ and $S(1) = I$. So, we have found a path within your set from $S$ to $I$. This shows that it is connected.

share|cite|improve this answer

answered Jul 18 at 21:21

amsmath

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
  – user9527
  Jul 18 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
  – amsmath
  Jul 18 at 21:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your nice answer.
  – user9527
  Jul 18 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
  – user9527
  Jul 19 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $GL(n,mathbb R)$ is not connected.
  – amsmath
  Jul 19 at 20:20
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Your condition $(S^-1AS)_cdot,1 = a_1$ is equivalent to $(SA-AS)e_1 = 0$.
Let $S$ be a matrix in your set and let $gamma : [0,1]tomathbb C$ be a path such that $gamma(0) = 0$, $gamma(1) = 1$, and $det(S + gamma(t)(I-S))neq 0$. It is now easy to see that $S(t) = S + gamma(t)(I-S)$ is in your set for each $tin [0,1]$ with $S(0) = S$ and $S(1) = I$. So, we have found a path within your set from $S$ to $I$. This shows that it is connected.

share|cite|improve this answer

answered Jul 18 at 21:21

amsmath

1,613114

Your condition $(S^-1AS)_cdot,1 = a_1$ is equivalent to $(SA-AS)e_1 = 0$.
Let $S$ be a matrix in your set and let $gamma : [0,1]tomathbb C$ be a path such that $gamma(0) = 0$, $gamma(1) = 1$, and $det(S + gamma(t)(I-S))neq 0$. It is now easy to see that $S(t) = S + gamma(t)(I-S)$ is in your set for each $tin [0,1]$ with $S(0) = S$ and $S(1) = I$. So, we have found a path within your set from $S$ to $I$. This shows that it is connected.

share|cite|improve this answer

answered Jul 18 at 21:21

amsmath

1,613114

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 21:21

amsmath

1,613114

answered Jul 18 at 21:21

amsmath

1,613114

answered Jul 18 at 21:21

amsmath

1,613114

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
  – user9527
  Jul 18 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
  – amsmath
  Jul 18 at 21:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your nice answer.
  – user9527
  Jul 18 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
  – user9527
  Jul 19 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $GL(n,mathbb R)$ is not connected.
  – amsmath
  Jul 19 at 20:20
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
  – user9527
  Jul 18 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
  – amsmath
  Jul 18 at 21:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your nice answer.
  – user9527
  Jul 18 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
  – user9527
  Jul 19 at 20:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $GL(n,mathbb R)$ is not connected.
  – amsmath
  Jul 19 at 20:20
  

  
  
  
  

May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
– user9527
Jul 18 at 21:34

May I ask why $gamma(t)$ exists such that $det(S + gamma(t)(I-S)) neq 0$?
– user9527
Jul 18 at 21:34

@iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
– amsmath
Jul 18 at 21:47

@iris2017 Sure, you may. Let $Z$ be the set of all $zinmathbb C$ such that $-z/(1-z)$ is an eigenvalue of $S$. Obviously, the set $Z$ is a finite set. Now, choose $gamma : [0,1]tomathbb C$ continuous such that $gamma(0) = 0$, $gamma(1) = 1$ and $gamma(t)notin Z$ for each $tin [0,1]$. This is obviously possible. Then$$S + gamma(t)(I-S) = (1-gamma(t))S + gamma(t)I = (1-gamma(t))left(S + fracgamma(t)1-gamma(t)Iright),$$is invertible for any $tin [0,1]$.
– amsmath
Jul 18 at 21:47

Thanks for your nice answer.
– user9527
Jul 18 at 22:04

Thanks for your nice answer.
– user9527
Jul 18 at 22:04

Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
– user9527
Jul 19 at 20:19

Could you comment on the case if we replace the filed $mathbb C$ by $mathbb R$? Thanks.
– user9527
Jul 19 at 20:19

$GL(n,mathbb R)$ is not connected.
– amsmath
Jul 19 at 20:20

$GL(n,mathbb R)$ is not connected.
– amsmath
Jul 19 at 20:20

 | 
show 2 more comments

 

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