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Prove that if and only if $(A,B)$ is controllable, then $(A−BK,B)$ is also controllable [closed]

0

$begingroup$

Given $A in mathbb R^ntimes n$, $B in mathbb R^ntimes 1$ and $K in mathbb R^1times n$, prove that

$(A,B)$ is controllable $Leftrightarrow$ $(A−BK,B)$ is controllable.

Any help would be greatly appreciated.

matrix-rank optimal-control linear-control

share|cite|improve this question

edited Mar 31 at 0:29

idriskameni

751321

asked Mar 30 at 22:51

AthosAthos

62

$endgroup$

closed as off-topic by Leucippus, Eevee Trainer, Cesareo, Shailesh, Paul Frost Mar 31 at 9:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Eevee Trainer, Cesareo, Shailesh, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your version of controllable? In terms of $A,B$ or in terms of dynamical systems?
  $endgroup$
  – copper.hat
  Mar 30 at 23:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Define how controllability is determined by the rank of the controllability matrix for the matrices (A, b) by C=[b, Ab…A^(n-1)b]. If C is full rank, then (A, b) is controllable.
  $endgroup$
  – Athos
  Mar 30 at 23:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For Trump's sake, give the person a chance. Slow down on the close votes.
  $endgroup$
  – copper.hat
  Mar 30 at 23:41
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Given $A in mathbb R^ntimes n$, $B in mathbb R^ntimes 1$ and $K in mathbb R^1times n$, prove that

$(A,B)$ is controllable $Leftrightarrow$ $(A−BK,B)$ is controllable.

Any help would be greatly appreciated.

matrix-rank optimal-control linear-control

share|cite|improve this question

edited Mar 31 at 0:29

idriskameni

751321

asked Mar 30 at 22:51

AthosAthos

62

$endgroup$

closed as off-topic by Leucippus, Eevee Trainer, Cesareo, Shailesh, Paul Frost Mar 31 at 9:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Eevee Trainer, Cesareo, Shailesh, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your version of controllable? In terms of $A,B$ or in terms of dynamical systems?
  $endgroup$
  – copper.hat
  Mar 30 at 23:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Define how controllability is determined by the rank of the controllability matrix for the matrices (A, b) by C=[b, Ab…A^(n-1)b]. If C is full rank, then (A, b) is controllable.
  $endgroup$
  – Athos
  Mar 30 at 23:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For Trump's sake, give the person a chance. Slow down on the close votes.
  $endgroup$
  – copper.hat
  Mar 30 at 23:41
  
  

  
  
  
  

add a comment | 

$begingroup$

Given $A in mathbb R^ntimes n$, $B in mathbb R^ntimes 1$ and $K in mathbb R^1times n$, prove that

$(A,B)$ is controllable $Leftrightarrow$ $(A−BK,B)$ is controllable.

Any help would be greatly appreciated.

matrix-rank optimal-control linear-control

share|cite|improve this question

edited Mar 31 at 0:29

idriskameni

751321

asked Mar 30 at 22:51

AthosAthos

62

$endgroup$

share|cite|improve this question

edited Mar 31 at 0:29

idriskameni

751321

asked Mar 30 at 22:51

AthosAthos

62

share|cite|improve this question

edited Mar 31 at 0:29

idriskameni

751321

asked Mar 30 at 22:51

AthosAthos

62

edited Mar 31 at 0:29

idriskameni

751321

edited Mar 31 at 0:29

idriskameni

751321

asked Mar 30 at 22:51

AthosAthos

62

asked Mar 30 at 22:51

AthosAthos

62

1 Answer
1

active

oldest

votes

1

$begingroup$

One criteria for controllability is (Popov Belevitch Hautus criteria)
$operatornamerk beginbmatrix lambda I -A & B endbmatrix = n$ for all $lambda$.

Suppose $(A-BK,B)$ is not controllable, then there is some $v$ such that
$v^* (lambda I -A+BK) = 0, v^* B = 0$
from which we get that $v^*(lambda I -A) =0, v^* B = 0$ hence $(A,B)$ is not controllable.

In a similar manner, we see that if $(A,B)$ is not controllable, then $(A-BK,B)$ is not controllable.

share|cite|improve this answer

edited Mar 30 at 23:51

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

1

$begingroup$

One criteria for controllability is (Popov Belevitch Hautus criteria)
$operatornamerk beginbmatrix lambda I -A & B endbmatrix = n$ for all $lambda$.

Suppose $(A-BK,B)$ is not controllable, then there is some $v$ such that
$v^* (lambda I -A+BK) = 0, v^* B = 0$
from which we get that $v^*(lambda I -A) =0, v^* B = 0$ hence $(A,B)$ is not controllable.

In a similar manner, we see that if $(A,B)$ is not controllable, then $(A-BK,B)$ is not controllable.

share|cite|improve this answer

edited Mar 30 at 23:51

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

1

$begingroup$

One criteria for controllability is (Popov Belevitch Hautus criteria)
$operatornamerk beginbmatrix lambda I -A & B endbmatrix = n$ for all $lambda$.

Suppose $(A-BK,B)$ is not controllable, then there is some $v$ such that
$v^* (lambda I -A+BK) = 0, v^* B = 0$
from which we get that $v^*(lambda I -A) =0, v^* B = 0$ hence $(A,B)$ is not controllable.

In a similar manner, we see that if $(A,B)$ is not controllable, then $(A-BK,B)$ is not controllable.

share|cite|improve this answer

edited Mar 30 at 23:51

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

$begingroup$

One criteria for controllability is (Popov Belevitch Hautus criteria)
$operatornamerk beginbmatrix lambda I -A & B endbmatrix = n$ for all $lambda$.

Suppose $(A-BK,B)$ is not controllable, then there is some $v$ such that
$v^* (lambda I -A+BK) = 0, v^* B = 0$
from which we get that $v^*(lambda I -A) =0, v^* B = 0$ hence $(A,B)$ is not controllable.

In a similar manner, we see that if $(A,B)$ is not controllable, then $(A-BK,B)$ is not controllable.

share|cite|improve this answer

edited Mar 30 at 23:51

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161

$endgroup$

share|cite|improve this answer

edited Mar 30 at 23:51

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161

answered Mar 30 at 23:22

copper.hatcopper.hat

128k561161