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If the spectral radius of a matrix is less than 1, can we conclude...
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If the spectral radius of a matrix is less than 1, can we conclude...
Dear Omran Kouba What if one eigenvalue has a positive real part regardless of the spectral selected norm? Do you think the system preserves stability? Regards
Dear Omran Kouba What if one eigenvalue has a positive real part regardless of the spectral selected norm? Do you think the system preserves stability? Regards
Dear Omran Kouba, It is a confusing criteria for stability. We are familiar with the first and second lyapunov approach and some analog methods such as the hurwitz criteria to guarantee the dynamical system stability of the equilibrium states. All based on the sign analysis of the eigenvalues and not their magnitude! Anyway, I appreciate if you mention some references that discuss and define this type of stability. Best regards
Dear Omran Kouba, It is a confusing criteria for stability. We are familiar with the first and second lyapunov approach and some analog methods such as the hurwitz criteria to guarantee the dynamical system stability of the equilibrium states. All based on the sign analysis of the eigenvalues and not their magnitude! Anyway, I appreciate if you mention some references that discuss and define this type of stability. Best regards
Om Prakash Yadav The answer to your question is Yes, because there is a norm N(.) on R2 such that the associated norm N' satisfies N'(A)<1. Take for example N(x)=Σ0≤kck||Akx||1 where ρ(A)<c<1 (for instance c=0.8.) with this choice N'(A)≤ c.
Om Prakash Yadav The answer to your question is Yes, because there is a norm N(.) on R2 such that the associated norm N' satisfies N'(A)<1. Take for example N(x)=Σ0≤kck||Akx||1 where ρ(A)with this choice N'(A)≤ c.
Dear Dr. Omran Kouba , Thank you very much for help. I would appreciate if you could add some insights for the forllowing problem, Suppose I have a matrix A=[0.4 0.8; 0 0.7]. Then 0.7=ρ(A)<1, but the infinity norm or the norm-1 are >1, 1.2 and 1.5 respectively, to be precise. May I conclude the iterative scheme x_k=Ax_{k-1} will be stable. Thanks for your time.
Dear Dr. Omran Kouba , Thank you very much for help. I would appreciate if you could add some insights for the forllowing problem, Suppose I have a matrix A=[0.4 0.8; 0 0.7]. Then 0.7=ρ(A)<1, but the infinity norm or the norm-1 are >1, 1.2 and 1.5 respectively, to be precise. May I conclude the iterative scheme x_k=Ax_{k-1} will be stable. Thanks for your time.
Dear Issam Kaddoura For discrete time systems stability depends on the magnitude of the eigenvalues of A , not the sign of the real part. Eigenvalues inside the unit circle = stability. Regards.
Dear Issam Kaddoura For discrete time systems stability depends on the magnitude of the eigenvalues of A , not the sign of the real part. Eigenvalues inside the unit circle = stability. Regards.
(1) Since the space Mn(R) of nxn real matrices is finite dimensional all norms on this space are equivalent. (2) If || . || is a norm on Rn then we have an associated matrix norm || . ||’ by setting ||A||’= sup{||Ax||: ||x||=1}. Not all norms on Mn(R) are associated to some norm on Rn. For example the Frobenius norm is not an associated one. (3) If || . ||’ is an associated norm on Mn(R) then ρ(A) ≤||A||’ and we do not have equality in general, however for every matrix A and every 0<ε<1 there exists an associated norm || . ||’ such that (1-ε) ||A||’ ≤ ρ(A) ≤||A||’. (4) In conclusion, If for some associated norm || .||’ we have ||A||’<1 then ρ(A)<1 and conversely if ρ(A)<1 then there exists an associated norm || .||’ such that ||A||’<1. So the following statements are equivalent
limk→∞ Ak = 0.
ρ(A)<1.
There exists an associated norm || . || such that ||A||‘<1.
(1) Since the space Mn(R) of nxn real matrices is finite dimensional all norms on this space are equivalent. (2) If || . || is a norm on Rn then we have an associated matrix norm || . ||’ by setting ||A||’= sup{||Ax||: ||x||=1}. Not all norms on Mn(R) are associated to some norm on Rn. For example the Frobenius norm is not an associated one. (3) If || . ||’ is an associated norm on Mn(R) then ρ(A) ≤||A||’ and we do not have equality in general, however for every matrix A and every 0<ε<1 there exists an associated norm || . ||’ such that (1-ε) ||A||’ ≤ ρ(A) ≤||A||’. (4) In conclusion, If for some associated norm || .||’ we have ||A||’<1 then ρ(A)<1 and conversely if ρ(A)<1 then there exists an associated norm || .||’ such that ||A||’<1. So the following statements are equivalent
limk→∞ Ak = 0.
ρ(A)<1.
There exists an associated norm || . || such that ||A||‘<1.
Dear Omran Kouba
What if one eigenvalue has a positive real part regardless of the spectral selected norm? Do you think the system preserves stability?
Regards
Dear Omran Kouba
What if one eigenvalue has a positive real part regardless of the spectral selected norm? Do you think the system preserves stability?
Regards
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Much appreciated Prof. Kouba, Omran Kouba
Much appreciated Prof. Kouba, Omran Kouba
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Dear Omran Kouba,
It is a confusing criteria for stability.
We are familiar with the first and second lyapunov approach and some analog methods such as the hurwitz criteria to guarantee the dynamical system stability of the equilibrium states. All based on the sign analysis of the eigenvalues and not their magnitude!
Anyway, I appreciate if you mention some references that discuss and define this type of stability.
Best regards
Dear Omran Kouba,
It is a confusing criteria for stability.
We are familiar with the first and second lyapunov approach and some analog methods such as the hurwitz criteria to guarantee the dynamical system stability of the equilibrium states. All based on the sign analysis of the eigenvalues and not their magnitude!
Anyway, I appreciate if you mention some references that discuss and define this type of stability.
Best regards
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Om Prakash Yadav
The answer to your question is Yes, because there is a norm N(.) on R2 such that the associated norm N' satisfies N'(A)<1.
Take for example N(x)=Σ0≤kck||Akx||1 where ρ(A)<c<1 (for instance c=0.8.)
with this choice N'(A)≤ c.
Om Prakash Yadav
The answer to your question is Yes, because there is a norm N(.) on R2 such that the associated norm N' satisfies N'(A)<1.
Take for example N(x)=Σ0≤kck||Akx||1 where ρ(A)
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Dear Dr. Omran Kouba , Thank you very much for help. I would appreciate if you could add some insights for the forllowing problem,
Suppose I have a matrix A=[0.4 0.8; 0 0.7]. Then 0.7=ρ(A)<1, but the infinity norm or the norm-1 are >1, 1.2 and 1.5 respectively, to be precise. May I conclude the iterative scheme x_k=Ax_{k-1} will be stable.
Thanks for your time.
Dear Dr. Omran Kouba , Thank you very much for help. I would appreciate if you could add some insights for the forllowing problem,
Suppose I have a matrix A=[0.4 0.8; 0 0.7]. Then 0.7=ρ(A)<1, but the infinity norm or the norm-1 are >1, 1.2 and 1.5 respectively, to be precise. May I conclude the iterative scheme x_k=Ax_{k-1} will be stable.
Thanks for your time.
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Dear Issam Kaddoura
Here is a tutorial on discrete time linear systems.
https://homes.esat.kuleuven.be/~maapc/static/files/SYSTHEORY/Slides/Lecture6/Lecture6-Discrete%20time%20system%20properties.pdf
You can also consult this
http://www.control.tu-berlin.de/images/0/07/DCS_ADTS_2x3.pdf
Dear Issam Kaddoura
Here is a tutorial on discrete time linear systems.
https://homes.esat.kuleuven.be/~maapc/static/files/SYSTHEORY/Slides/Lecture6/Lecture6-Discrete%20time%20system%20properties.pdf
You can also consult this
http://www.control.tu-berlin.de/images/0/07/DCS_ADTS_2x3.pdf
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Dear Dr. Issam Kaddoura I am having a matrix of tridiagonal type and studying the stability of an iterative scheme x_{k+1}=Ax_k.
Thanks for your time.
Dear Dr. Issam Kaddoura I am having a matrix of tridiagonal type and studying the stability of an iterative scheme x_{k+1}=Ax_k.
Thanks for your time.
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Dear Issam Kaddoura
For discrete time systems stability depends on the magnitude of the eigenvalues of A , not the sign of the real part. Eigenvalues inside the unit circle = stability.
Regards.
Dear Issam Kaddoura
For discrete time systems stability depends on the magnitude of the eigenvalues of A , not the sign of the real part. Eigenvalues inside the unit circle = stability.
Regards.
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Thanks a lot, Sir.
Thanks a lot, Sir.
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(1) Since the space Mn(R) of nxn real matrices is finite dimensional all norms on this space are equivalent.
(2) If || . || is a norm on Rn then we have an associated matrix norm || . ||’ by setting ||A||’= sup{||Ax||: ||x||=1}. Not all norms on Mn(R) are associated to some norm on Rn. For example the Frobenius norm is not an associated one.
(3) If || . ||’ is an associated norm on Mn(R) then ρ(A) ≤ ||A||’ and we do not have equality in general, however for every matrix A and every 0<ε<1 there exists an associated norm || . ||’ such that (1-ε) ||A||’ ≤ ρ(A) ≤ ||A||’.
(4) In conclusion, If for some associated norm || .||’ we have ||A||’<1 then ρ(A)<1 and conversely if ρ(A)<1 then there exists an associated norm || .||’ such that ||A||’<1. So the following statements are equivalent
(1) Since the space Mn(R) of nxn real matrices is finite dimensional all norms on this space are equivalent.
(2) If || . || is a norm on Rn then we have an associated matrix norm || . ||’ by setting ||A||’= sup{||Ax||: ||x||=1}. Not all norms on Mn(R) are associated to some norm on Rn. For example the Frobenius norm is not an associated one.
(3) If || . ||’ is an associated norm on Mn(R) then ρ(A) ≤ ||A||’ and we do not have equality in general, however for every matrix A and every 0<ε<1 there exists an associated norm || . ||’ such that (1-ε) ||A||’ ≤ ρ(A) ≤ ||A||’.
(4) In conclusion, If for some associated norm || .||’ we have ||A||’<1 then ρ(A)<1 and conversely if ρ(A)<1 then there exists an associated norm || .||’ such that ||A||’<1. So the following statements are equivalent
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