Markov matrices with <=1 absorbing states have all but one eigen values <1?
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I found plenty of proofs online that a Markov matrix will have all its eigen values of modulus <=1. In the book on Introduction to Matrix analysis by Bellman, he also shows in section 8 of chapter 14 that for positive Markov matrices, the only eigen value of magnitude 1 can be 1 itself (and not -1). I beleive positive Markov matrices imply all their elements are positive and so exclude matrices with absorbing states.
But, what I've observed is that for all such matrices I've come across, there is just one eigen value, 1 and all other eigen values are <1 in magnitude.
In fact, I've observed this for Markov matrices with just one absorbing state as well.
Is there a way to prove or refute (with counterexamples or otherwise) the above two claims?
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
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add a comment |
$begingroup$
I found plenty of proofs online that a Markov matrix will have all its eigen values of modulus <=1. In the book on Introduction to Matrix analysis by Bellman, he also shows in section 8 of chapter 14 that for positive Markov matrices, the only eigen value of magnitude 1 can be 1 itself (and not -1). I beleive positive Markov matrices imply all their elements are positive and so exclude matrices with absorbing states.
But, what I've observed is that for all such matrices I've come across, there is just one eigen value, 1 and all other eigen values are <1 in magnitude.
In fact, I've observed this for Markov matrices with just one absorbing state as well.
Is there a way to prove or refute (with counterexamples or otherwise) the above two claims?
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
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1
$begingroup$
See numbertheory.org/courses/MP274/markov.pdf for a start.
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– amd
Jan 17 at 0:46
add a comment |
$begingroup$
I found plenty of proofs online that a Markov matrix will have all its eigen values of modulus <=1. In the book on Introduction to Matrix analysis by Bellman, he also shows in section 8 of chapter 14 that for positive Markov matrices, the only eigen value of magnitude 1 can be 1 itself (and not -1). I beleive positive Markov matrices imply all their elements are positive and so exclude matrices with absorbing states.
But, what I've observed is that for all such matrices I've come across, there is just one eigen value, 1 and all other eigen values are <1 in magnitude.
In fact, I've observed this for Markov matrices with just one absorbing state as well.
Is there a way to prove or refute (with counterexamples or otherwise) the above two claims?
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
I found plenty of proofs online that a Markov matrix will have all its eigen values of modulus <=1. In the book on Introduction to Matrix analysis by Bellman, he also shows in section 8 of chapter 14 that for positive Markov matrices, the only eigen value of magnitude 1 can be 1 itself (and not -1). I beleive positive Markov matrices imply all their elements are positive and so exclude matrices with absorbing states.
But, what I've observed is that for all such matrices I've come across, there is just one eigen value, 1 and all other eigen values are <1 in magnitude.
In fact, I've observed this for Markov matrices with just one absorbing state as well.
Is there a way to prove or refute (with counterexamples or otherwise) the above two claims?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
edited Jan 16 at 6:33
Rohit Pandey
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
asked Jan 16 at 6:25
Rohit PandeyRohit Pandey
1,6581024
1,6581024
1
$begingroup$
See numbertheory.org/courses/MP274/markov.pdf for a start.
$endgroup$
– amd
Jan 17 at 0:46
add a comment |
1
$begingroup$
See numbertheory.org/courses/MP274/markov.pdf for a start.
$endgroup$
– amd
Jan 17 at 0:46
1
1
$begingroup$
See numbertheory.org/courses/MP274/markov.pdf for a start.
$endgroup$
– amd
Jan 17 at 0:46
$begingroup$
See numbertheory.org/courses/MP274/markov.pdf for a start.
$endgroup$
– amd
Jan 17 at 0:46
add a comment |
1 Answer
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To make the question answered I note that a Markov matrix $$begin{pmatrix}
1 & 0 & 0\
0 &0 & 1\
0 &1 & 0end{pmatrix}$$ corresponding to a chain with just one absorbing state should refute the claim, because it has eigenvalues $1$ and $-1$
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
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$begingroup$
To make the question answered I note that a Markov matrix $$begin{pmatrix}
1 & 0 & 0\
0 &0 & 1\
0 &1 & 0end{pmatrix}$$ corresponding to a chain with just one absorbing state should refute the claim, because it has eigenvalues $1$ and $-1$
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
$begingroup$
To make the question answered I note that a Markov matrix $$begin{pmatrix}
1 & 0 & 0\
0 &0 & 1\
0 &1 & 0end{pmatrix}$$ corresponding to a chain with just one absorbing state should refute the claim, because it has eigenvalues $1$ and $-1$
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
$begingroup$
To make the question answered I note that a Markov matrix $$begin{pmatrix}
1 & 0 & 0\
0 &0 & 1\
0 &1 & 0end{pmatrix}$$ corresponding to a chain with just one absorbing state should refute the claim, because it has eigenvalues $1$ and $-1$
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
To make the question answered I note that a Markov matrix $$begin{pmatrix}
1 & 0 & 0\
0 &0 & 1\
0 &1 & 0end{pmatrix}$$ corresponding to a chain with just one absorbing state should refute the claim, because it has eigenvalues $1$ and $-1$
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
answered Jan 19 at 9:38
Alex RavskyAlex Ravsky
42.7k32483
42.7k32483
add a comment |
add a comment |
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1
$begingroup$
See numbertheory.org/courses/MP274/markov.pdf for a start.
$endgroup$
– amd
Jan 17 at 0:46