Unique right eigenvector for row stochastic matrix
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Suppose $P$ is a row-stochastic matrix. Then $1$ is a right-eigenvalue of $P$, and any constant vector $v$ will satisfy $Pv=v$. When is it the case that only constant vectors solve $Pv=v$? In particular, I am wondering if this is the case if if $P$ is irreducible and aperiodic.
linear-algebra markov-chains
asked Dec 19 '18 at 23:22
MikeMike
368110
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add a comment |
$begingroup$
Suppose $P$ is a row-stochastic matrix. Then $1$ is a right-eigenvalue of $P$, and any constant vector $v$ will satisfy $Pv=v$. When is it the case that only constant vectors solve $Pv=v$? In particular, I am wondering if this is the case if if $P$ is irreducible and aperiodic.
linear-algebra markov-chains
asked Dec 19 '18 at 23:22
MikeMike
368110
$endgroup$
add a comment |
$begingroup$
Suppose $P$ is a row-stochastic matrix. Then $1$ is a right-eigenvalue of $P$, and any constant vector $v$ will satisfy $Pv=v$. When is it the case that only constant vectors solve $Pv=v$? In particular, I am wondering if this is the case if if $P$ is irreducible and aperiodic.
linear-algebra markov-chains
asked Dec 19 '18 at 23:22
MikeMike
368110
$endgroup$
Suppose $P$ is a row-stochastic matrix. Then $1$ is a right-eigenvalue of $P$, and any constant vector $v$ will satisfy $Pv=v$. When is it the case that only constant vectors solve $Pv=v$? In particular, I am wondering if this is the case if if $P$ is irreducible and aperiodic.
linear-algebra markov-chains
linear-algebra markov-chains
asked Dec 19 '18 at 23:22
MikeMike
368110
asked Dec 19 '18 at 23:22
MikeMike
368110
asked Dec 19 '18 at 23:22
MikeMike
368110
asked Dec 19 '18 at 23:22
MikeMike
368110
asked Dec 19 '18 at 23:22
MikeMike
368110
368110
add a comment |
add a comment |
1 Answer
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The condition that $P$ is irreducible is enough to guarantee that property.
We need to show that $Pv=v$ implies $v_1=ldots=v_n$. Let $v_p$ be the smallest component of $v$, so that $v_i ge v_p$ for all $i$. Let $U subseteq {1,ldots,n}$ be the set of those indices for which $v_i=v_p$. Note that $Uneq emptyset$ since $p in U$. If $m in U$, then
$$
(P v)_m = v_m = v_p.
$$
Now, because $P$ is row-stochastic, the left hand side of the above equation is a convex combination of the components of $v$, while the right hand side is a minimum-value component of $v$. The only way that a convex combination of elements can yield a value equal to the minimum element is when all elements with a positive coefficient (here, all $v_k$ such that $P_{km}>0$) are equal to the minimum. Thus, $m in U$ implies $k in U$ whenever $P_{km}>0$. Since $P$ is irreducible, this implies $U={1,ldots,n}$, hence $v_1=ldots=v_n$.
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
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$begingroup$
The condition that $P$ is irreducible is enough to guarantee that property.
We need to show that $Pv=v$ implies $v_1=ldots=v_n$. Let $v_p$ be the smallest component of $v$, so that $v_i ge v_p$ for all $i$. Let $U subseteq {1,ldots,n}$ be the set of those indices for which $v_i=v_p$. Note that $Uneq emptyset$ since $p in U$. If $m in U$, then
$$
(P v)_m = v_m = v_p.
$$
Now, because $P$ is row-stochastic, the left hand side of the above equation is a convex combination of the components of $v$, while the right hand side is a minimum-value component of $v$. The only way that a convex combination of elements can yield a value equal to the minimum element is when all elements with a positive coefficient (here, all $v_k$ such that $P_{km}>0$) are equal to the minimum. Thus, $m in U$ implies $k in U$ whenever $P_{km}>0$. Since $P$ is irreducible, this implies $U={1,ldots,n}$, hence $v_1=ldots=v_n$.
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
$endgroup$
add a comment |
$begingroup$
The condition that $P$ is irreducible is enough to guarantee that property.
We need to show that $Pv=v$ implies $v_1=ldots=v_n$. Let $v_p$ be the smallest component of $v$, so that $v_i ge v_p$ for all $i$. Let $U subseteq {1,ldots,n}$ be the set of those indices for which $v_i=v_p$. Note that $Uneq emptyset$ since $p in U$. If $m in U$, then
$$
(P v)_m = v_m = v_p.
$$
Now, because $P$ is row-stochastic, the left hand side of the above equation is a convex combination of the components of $v$, while the right hand side is a minimum-value component of $v$. The only way that a convex combination of elements can yield a value equal to the minimum element is when all elements with a positive coefficient (here, all $v_k$ such that $P_{km}>0$) are equal to the minimum. Thus, $m in U$ implies $k in U$ whenever $P_{km}>0$. Since $P$ is irreducible, this implies $U={1,ldots,n}$, hence $v_1=ldots=v_n$.
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
$endgroup$
add a comment |
$begingroup$
The condition that $P$ is irreducible is enough to guarantee that property.
We need to show that $Pv=v$ implies $v_1=ldots=v_n$. Let $v_p$ be the smallest component of $v$, so that $v_i ge v_p$ for all $i$. Let $U subseteq {1,ldots,n}$ be the set of those indices for which $v_i=v_p$. Note that $Uneq emptyset$ since $p in U$. If $m in U$, then
$$
(P v)_m = v_m = v_p.
$$
Now, because $P$ is row-stochastic, the left hand side of the above equation is a convex combination of the components of $v$, while the right hand side is a minimum-value component of $v$. The only way that a convex combination of elements can yield a value equal to the minimum element is when all elements with a positive coefficient (here, all $v_k$ such that $P_{km}>0$) are equal to the minimum. Thus, $m in U$ implies $k in U$ whenever $P_{km}>0$. Since $P$ is irreducible, this implies $U={1,ldots,n}$, hence $v_1=ldots=v_n$.
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
$endgroup$
The condition that $P$ is irreducible is enough to guarantee that property.
We need to show that $Pv=v$ implies $v_1=ldots=v_n$. Let $v_p$ be the smallest component of $v$, so that $v_i ge v_p$ for all $i$. Let $U subseteq {1,ldots,n}$ be the set of those indices for which $v_i=v_p$. Note that $Uneq emptyset$ since $p in U$. If $m in U$, then
$$
(P v)_m = v_m = v_p.
$$
Now, because $P$ is row-stochastic, the left hand side of the above equation is a convex combination of the components of $v$, while the right hand side is a minimum-value component of $v$. The only way that a convex combination of elements can yield a value equal to the minimum element is when all elements with a positive coefficient (here, all $v_k$ such that $P_{km}>0$) are equal to the minimum. Thus, $m in U$ implies $k in U$ whenever $P_{km}>0$. Since $P$ is irreducible, this implies $U={1,ldots,n}$, hence $v_1=ldots=v_n$.
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
answered Dec 19 '18 at 23:51
VincenzoVincenzo
21816
21816
add a comment |
add a comment |
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