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Is it possible to upper bound this family of matrices in operator norm?

2

$begingroup$

Let

$$mathcal E = _2$$

where $A_0 $ is some fixed matrix and $|cdot|_2$ denotes the induced $2$-norm. We also have for every $A in mathcal E$, $rho(A)< 1$ where $rho(cdot)$ denotes the spectral radius and $rho(A_0) < 1$. Is it possible to give an upper bound $C$ in terms of $|A_0|_2$ such that $|(I-A)^-1|_2 le C$ for all $A in mathcal E$?

linear-algebra matrices spectral-radius matrix-norms spectral-norm

share|cite|improve this question

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Notably, $mathcal E$ is compact. So, there necessarily exists a stronger upper bound $r < 1$ such that $rho(A) leq r$ for all $A in mathcal E$.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does $|A|_2$ denote the Frobenius norm or the induced 2-norm?
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:41
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: It is induced $2$-norm.
  $endgroup$
  – user1101010
  Jun 21 '18 at 22:42
  

  
  
  
  

add a comment | 

2

$begingroup$

Let

$$mathcal E = _2$$

where $A_0 $ is some fixed matrix and $|cdot|_2$ denotes the induced $2$-norm. We also have for every $A in mathcal E$, $rho(A)< 1$ where $rho(cdot)$ denotes the spectral radius and $rho(A_0) < 1$. Is it possible to give an upper bound $C$ in terms of $|A_0|_2$ such that $|(I-A)^-1|_2 le C$ for all $A in mathcal E$?

linear-algebra matrices spectral-radius matrix-norms spectral-norm

share|cite|improve this question

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Notably, $mathcal E$ is compact. So, there necessarily exists a stronger upper bound $r < 1$ such that $rho(A) leq r$ for all $A in mathcal E$.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does $|A|_2$ denote the Frobenius norm or the induced 2-norm?
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:41
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: It is induced $2$-norm.
  $endgroup$
  – user1101010
  Jun 21 '18 at 22:42
  

  
  
  
  

add a comment | 

$begingroup$

Let

$$mathcal E = _2$$

where $A_0 $ is some fixed matrix and $|cdot|_2$ denotes the induced $2$-norm. We also have for every $A in mathcal E$, $rho(A)< 1$ where $rho(cdot)$ denotes the spectral radius and $rho(A_0) < 1$. Is it possible to give an upper bound $C$ in terms of $|A_0|_2$ such that $|(I-A)^-1|_2 le C$ for all $A in mathcal E$?

linear-algebra matrices spectral-radius matrix-norms spectral-norm

share|cite|improve this question

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

$endgroup$

share|cite|improve this question

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

share|cite|improve this question

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

edited Mar 31 at 7:27

Rodrigo de Azevedo

13.2k41962

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

asked Jun 21 '18 at 22:24

user1101010user1101010

9011830

1 Answer
1

active

oldest

votes

1

$begingroup$

The answer is yes. The upper bound I come up with below is $frac1A_0$.

It doesn't seem like there's any need to consider the matrix $A_0$ itself. In the below, we will take $M = |A|_0 geq 0$, since no other information about $A_0$ will be used.

Suppose that $M geq 1$. Then $mathcal E$ includes the identity matrix, and so we fail to have $rho(A) < 1$.

Suppose that $M < 1$. We note that
$$
|(I-A)^-1|_2 = left|sum_k=0^infty A^kright|_2 leq
sum_k=0^infty left|A^kright|_2 leq sum_k=0^infty left|Aright|_2^k leq sum_k=0^infty M^k = frac11 - M
$$

share|cite|improve this answer

edited Jun 21 '18 at 22:52

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer simplifies to the simplest answer to the question: Yes! (Sorry, couldn't help myself.)
  $endgroup$
  – ulaff.net
  Jun 21 '18 at 22:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ulaff.net I agree that any answer to a well formed yes/no question should include either a yes or a no; I've edited accordingly.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: Thanks. I realized I formulated my question in a wrong way. Indeed, the family of interest is for the case $M > 1$ and the family $mathcal E = _2 le M text and rho(A) < 1$. Anyway, your answer does addresses the question I asked.
  $endgroup$
  – user1101010
  Jun 21 '18 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iris2017 you should make a new post with the correctly formulated question, if you haven't done so already
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iris2017 also, I'm fairly confident that you'll only get the upper bound you want if you define your family as $$ mathcal E = _2 leq M text and rho(A) leq r $$ for some fixed $r$ with $0 leq r < 1$. For this set, it is notable that an upper bound must exist since $mathcal E$ is compact and the inverse function is continuous.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 23:04
  
  

  
  
  
  

 | 
show 1 more comment

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1

$begingroup$

The answer is yes. The upper bound I come up with below is $frac1A_0$.

It doesn't seem like there's any need to consider the matrix $A_0$ itself. In the below, we will take $M = |A|_0 geq 0$, since no other information about $A_0$ will be used.

Suppose that $M geq 1$. Then $mathcal E$ includes the identity matrix, and so we fail to have $rho(A) < 1$.

Suppose that $M < 1$. We note that
$$
|(I-A)^-1|_2 = left|sum_k=0^infty A^kright|_2 leq
sum_k=0^infty left|A^kright|_2 leq sum_k=0^infty left|Aright|_2^k leq sum_k=0^infty M^k = frac11 - M
$$

share|cite|improve this answer

edited Jun 21 '18 at 22:52

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer simplifies to the simplest answer to the question: Yes! (Sorry, couldn't help myself.)
  $endgroup$
  – ulaff.net
  Jun 21 '18 at 22:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ulaff.net I agree that any answer to a well formed yes/no question should include either a yes or a no; I've edited accordingly.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: Thanks. I realized I formulated my question in a wrong way. Indeed, the family of interest is for the case $M > 1$ and the family $mathcal E = _2 le M text and rho(A) < 1$. Anyway, your answer does addresses the question I asked.
  $endgroup$
  – user1101010
  Jun 21 '18 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iris2017 you should make a new post with the correctly formulated question, if you haven't done so already
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iris2017 also, I'm fairly confident that you'll only get the upper bound you want if you define your family as $$ mathcal E = _2 leq M text and rho(A) leq r $$ for some fixed $r$ with $0 leq r < 1$. For this set, it is notable that an upper bound must exist since $mathcal E$ is compact and the inverse function is continuous.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 23:04
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

The answer is yes. The upper bound I come up with below is $frac1A_0$.

It doesn't seem like there's any need to consider the matrix $A_0$ itself. In the below, we will take $M = |A|_0 geq 0$, since no other information about $A_0$ will be used.

Suppose that $M geq 1$. Then $mathcal E$ includes the identity matrix, and so we fail to have $rho(A) < 1$.

Suppose that $M < 1$. We note that
$$
|(I-A)^-1|_2 = left|sum_k=0^infty A^kright|_2 leq
sum_k=0^infty left|A^kright|_2 leq sum_k=0^infty left|Aright|_2^k leq sum_k=0^infty M^k = frac11 - M
$$

share|cite|improve this answer

edited Jun 21 '18 at 22:52

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer simplifies to the simplest answer to the question: Yes! (Sorry, couldn't help myself.)
  $endgroup$
  – ulaff.net
  Jun 21 '18 at 22:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ulaff.net I agree that any answer to a well formed yes/no question should include either a yes or a no; I've edited accordingly.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 22:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: Thanks. I realized I formulated my question in a wrong way. Indeed, the family of interest is for the case $M > 1$ and the family $mathcal E = _2 le M text and rho(A) < 1$. Anyway, your answer does addresses the question I asked.
  $endgroup$
  – user1101010
  Jun 21 '18 at 22:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iris2017 you should make a new post with the correctly formulated question, if you haven't done so already
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iris2017 also, I'm fairly confident that you'll only get the upper bound you want if you define your family as $$ mathcal E = _2 leq M text and rho(A) leq r $$ for some fixed $r$ with $0 leq r < 1$. For this set, it is notable that an upper bound must exist since $mathcal E$ is compact and the inverse function is continuous.
  $endgroup$
  – Omnomnomnom
  Jun 21 '18 at 23:04
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

The answer is yes. The upper bound I come up with below is $frac1A_0$.

It doesn't seem like there's any need to consider the matrix $A_0$ itself. In the below, we will take $M = |A|_0 geq 0$, since no other information about $A_0$ will be used.

Suppose that $M geq 1$. Then $mathcal E$ includes the identity matrix, and so we fail to have $rho(A) < 1$.

Suppose that $M < 1$. We note that
$$
|(I-A)^-1|_2 = left|sum_k=0^infty A^kright|_2 leq
sum_k=0^infty left|A^kright|_2 leq sum_k=0^infty left|Aright|_2^k leq sum_k=0^infty M^k = frac11 - M
$$

share|cite|improve this answer

edited Jun 21 '18 at 22:52

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188

$endgroup$

share|cite|improve this answer

edited Jun 21 '18 at 22:52

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188

answered Jun 21 '18 at 22:48

OmnomnomnomOmnomnomnom

129k794188