Spectral radius equal to 1 and convergence
The following theorem is well-known:
$$
lim_k A^k = 0 text{ if and only if } rho(A)<1
$$
(see wiki for context and proofs).
What if now $rho(A)=1$ and $lambdaneq -1$ for all $lambda in Spec(A)$. Then we also have convergence ? (not to $0$ but $A$ converges)
sequences-and-series matrices convergence eigenvalues-eigenvectors spectral-radius
asked Dec 6 at 16:33
Smilia
585516
add a comment |
The following theorem is well-known:
$$
lim_k A^k = 0 text{ if and only if } rho(A)<1
$$
(see wiki for context and proofs).
What if now $rho(A)=1$ and $lambdaneq -1$ for all $lambda in Spec(A)$. Then we also have convergence ? (not to $0$ but $A$ converges)
sequences-and-series matrices convergence eigenvalues-eigenvectors spectral-radius
asked Dec 6 at 16:33
Smilia
585516
add a comment |
The following theorem is well-known:
$$
lim_k A^k = 0 text{ if and only if } rho(A)<1
$$
(see wiki for context and proofs).
What if now $rho(A)=1$ and $lambdaneq -1$ for all $lambda in Spec(A)$. Then we also have convergence ? (not to $0$ but $A$ converges)
sequences-and-series matrices convergence eigenvalues-eigenvectors spectral-radius
asked Dec 6 at 16:33
Smilia
585516
The following theorem is well-known:
$$
lim_k A^k = 0 text{ if and only if } rho(A)<1
$$
(see wiki for context and proofs).
What if now $rho(A)=1$ and $lambdaneq -1$ for all $lambda in Spec(A)$. Then we also have convergence ? (not to $0$ but $A$ converges)
sequences-and-series matrices convergence eigenvalues-eigenvectors spectral-radius
sequences-and-series matrices convergence eigenvalues-eigenvectors spectral-radius
asked Dec 6 at 16:33
Smilia
585516
asked Dec 6 at 16:33
Smilia
585516
edited Dec 6 at 16:48
asked Dec 6 at 16:33
Smilia
585516
asked Dec 6 at 16:33
Smilia
585516
asked Dec 6 at 16:33
Smilia
585516
585516
add a comment |
add a comment |
1 Answer
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No. This already fails when all eigenvalues of $A$ are ones. E.g. $A^k=pmatrix{1&k\ 0&1}$ doesn't converge when $A=pmatrix{1&1\ 0&1}$.
When $|lambda|=1$ but $lambdane1$, we don't even have convergence of $lambda^k$.
answered Dec 6 at 17:31
user1551
71.2k566125
add a comment |
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1 Answer
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No. This already fails when all eigenvalues of $A$ are ones. E.g. $A^k=pmatrix{1&k\ 0&1}$ doesn't converge when $A=pmatrix{1&1\ 0&1}$.
When $|lambda|=1$ but $lambdane1$, we don't even have convergence of $lambda^k$.
answered Dec 6 at 17:31
user1551
71.2k566125
add a comment |
No. This already fails when all eigenvalues of $A$ are ones. E.g. $A^k=pmatrix{1&k\ 0&1}$ doesn't converge when $A=pmatrix{1&1\ 0&1}$.
When $|lambda|=1$ but $lambdane1$, we don't even have convergence of $lambda^k$.
answered Dec 6 at 17:31
user1551
71.2k566125
add a comment |
No. This already fails when all eigenvalues of $A$ are ones. E.g. $A^k=pmatrix{1&k\ 0&1}$ doesn't converge when $A=pmatrix{1&1\ 0&1}$.
When $|lambda|=1$ but $lambdane1$, we don't even have convergence of $lambda^k$.
answered Dec 6 at 17:31
user1551
71.2k566125
No. This already fails when all eigenvalues of $A$ are ones. E.g. $A^k=pmatrix{1&k\ 0&1}$ doesn't converge when $A=pmatrix{1&1\ 0&1}$.
When $|lambda|=1$ but $lambdane1$, we don't even have convergence of $lambda^k$.
answered Dec 6 at 17:31
user1551
71.2k566125
answered Dec 6 at 17:31
user1551
71.2k566125
answered Dec 6 at 17:31
user1551
71.2k566125
answered Dec 6 at 17:31
user1551
71.2k566125
71.2k566125
add a comment |
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