Is uniqueness of steady-state vector sufficient to regular transition matrix?
$begingroup$
If P is a transition matrix, then a steady-state vector for is a probability vector q such that $Pmathrm{q}=mathrm{q}$.
A transition matrix P is regular if some power $P^k$ contain only strictly positive entries.
We know that the steady-state vector is the eigenvector of P associated with the eignvalue 1.
So if rank(P-I)=Dim(P)-1, then the eigenvector of P which column sum is 1 (steady-state vector) will be unique.
Is uniqueness of steady-state vector sufficient to the regularness of P?
linear-algebra markov-chains
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
$endgroup$
add a comment |
$begingroup$
If P is a transition matrix, then a steady-state vector for is a probability vector q such that $Pmathrm{q}=mathrm{q}$.
A transition matrix P is regular if some power $P^k$ contain only strictly positive entries.
We know that the steady-state vector is the eigenvector of P associated with the eignvalue 1.
So if rank(P-I)=Dim(P)-1, then the eigenvector of P which column sum is 1 (steady-state vector) will be unique.
Is uniqueness of steady-state vector sufficient to the regularness of P?
linear-algebra markov-chains
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
$endgroup$
add a comment |
$begingroup$
If P is a transition matrix, then a steady-state vector for is a probability vector q such that $Pmathrm{q}=mathrm{q}$.
A transition matrix P is regular if some power $P^k$ contain only strictly positive entries.
We know that the steady-state vector is the eigenvector of P associated with the eignvalue 1.
So if rank(P-I)=Dim(P)-1, then the eigenvector of P which column sum is 1 (steady-state vector) will be unique.
Is uniqueness of steady-state vector sufficient to the regularness of P?
linear-algebra markov-chains
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
$endgroup$
If P is a transition matrix, then a steady-state vector for is a probability vector q such that $Pmathrm{q}=mathrm{q}$.
A transition matrix P is regular if some power $P^k$ contain only strictly positive entries.
We know that the steady-state vector is the eigenvector of P associated with the eignvalue 1.
So if rank(P-I)=Dim(P)-1, then the eigenvector of P which column sum is 1 (steady-state vector) will be unique.
Is uniqueness of steady-state vector sufficient to the regularness of P?
linear-algebra markov-chains
linear-algebra markov-chains
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
asked Nov 30 '18 at 18:38
Charles BaoCharles Bao
751919
751919
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Have already found a counterexample.
$$P_2=begin{pmatrix}
0 & 0.5 & 0.5 & 0\
0.5 & 0 & 0 & 0.5\
0.5 & 0 & 0 & 0.5\
0 & 0.5 & 0.5 & 0\
end{pmatrix}$$
with unique steady-state vector $(0.25,0.25,0.25,0.25)^T$
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Have already found a counterexample.
$$P_2=begin{pmatrix}
0 & 0.5 & 0.5 & 0\
0.5 & 0 & 0 & 0.5\
0.5 & 0 & 0 & 0.5\
0 & 0.5 & 0.5 & 0\
end{pmatrix}$$
with unique steady-state vector $(0.25,0.25,0.25,0.25)^T$
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
$endgroup$
add a comment |
$begingroup$
Have already found a counterexample.
$$P_2=begin{pmatrix}
0 & 0.5 & 0.5 & 0\
0.5 & 0 & 0 & 0.5\
0.5 & 0 & 0 & 0.5\
0 & 0.5 & 0.5 & 0\
end{pmatrix}$$
with unique steady-state vector $(0.25,0.25,0.25,0.25)^T$
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
$endgroup$
add a comment |
$begingroup$
Have already found a counterexample.
$$P_2=begin{pmatrix}
0 & 0.5 & 0.5 & 0\
0.5 & 0 & 0 & 0.5\
0.5 & 0 & 0 & 0.5\
0 & 0.5 & 0.5 & 0\
end{pmatrix}$$
with unique steady-state vector $(0.25,0.25,0.25,0.25)^T$
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
$endgroup$
Have already found a counterexample.
$$P_2=begin{pmatrix}
0 & 0.5 & 0.5 & 0\
0.5 & 0 & 0 & 0.5\
0.5 & 0 & 0 & 0.5\
0 & 0.5 & 0.5 & 0\
end{pmatrix}$$
with unique steady-state vector $(0.25,0.25,0.25,0.25)^T$
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
answered Nov 30 '18 at 19:10
Charles BaoCharles Bao
751919
751919
add a comment |
add a comment |
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