How to calculate row sums of a power of a matrix
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Let $P $ be an $ntimes n$ matrix whose row sums $=1$.Then how to calculate the row sums of $P^m$ where $m $ is a positive integer?
matrices soft-question
asked Oct 20 '14 at 3:17
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17.5k32494
add a comment |
up vote
2
down vote
favorite
Let $P $ be an $ntimes n$ matrix whose row sums $=1$.Then how to calculate the row sums of $P^m$ where $m $ is a positive integer?
matrices soft-question
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
Do you mean that the sum of each row is equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:21
What kind of entries does P have?i.e. real, complex, integers?
– shooting-squirrel
Oct 20 '14 at 3:21
entries are real and the row sums=1 means each row adds to 1
– Learnmore
Oct 20 '14 at 3:23
So all the values on a row, added up, equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:24
Yes all the values on a row, added up, equal to 1
– Learnmore
Oct 20 '14 at 3:43
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $P $ be an $ntimes n$ matrix whose row sums $=1$.Then how to calculate the row sums of $P^m$ where $m $ is a positive integer?
matrices soft-question
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
Let $P $ be an $ntimes n$ matrix whose row sums $=1$.Then how to calculate the row sums of $P^m$ where $m $ is a positive integer?
matrices soft-question
matrices soft-question
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
asked Oct 20 '14 at 3:17
Learnmore
17.5k32494
17.5k32494
Do you mean that the sum of each row is equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:21
What kind of entries does P have?i.e. real, complex, integers?
– shooting-squirrel
Oct 20 '14 at 3:21
entries are real and the row sums=1 means each row adds to 1
– Learnmore
Oct 20 '14 at 3:23
So all the values on a row, added up, equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:24
Yes all the values on a row, added up, equal to 1
– Learnmore
Oct 20 '14 at 3:43
add a comment |
Do you mean that the sum of each row is equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:21
What kind of entries does P have?i.e. real, complex, integers?
– shooting-squirrel
Oct 20 '14 at 3:21
entries are real and the row sums=1 means each row adds to 1
– Learnmore
Oct 20 '14 at 3:23
So all the values on a row, added up, equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:24
Yes all the values on a row, added up, equal to 1
– Learnmore
Oct 20 '14 at 3:43
Do you mean that the sum of each row is equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:21
Do you mean that the sum of each row is equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:21
What kind of entries does P have?i.e. real, complex, integers?
– shooting-squirrel
Oct 20 '14 at 3:21
What kind of entries does P have?i.e. real, complex, integers?
– shooting-squirrel
Oct 20 '14 at 3:21
entries are real and the row sums=1 means each row adds to 1
– Learnmore
Oct 20 '14 at 3:23
entries are real and the row sums=1 means each row adds to 1
– Learnmore
Oct 20 '14 at 3:23
So all the values on a row, added up, equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:24
So all the values on a row, added up, equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:24
Yes all the values on a row, added up, equal to 1
– Learnmore
Oct 20 '14 at 3:43
Yes all the values on a row, added up, equal to 1
– Learnmore
Oct 20 '14 at 3:43
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The row sums of any power of $P$ are always 1. To see this, write $[P]_{ij}:=p_{ij}$. Then
$$[P^2]_{ij}=sum_{k}^np_{ik}p_{kj}$$
and
$$sum_{j=1}^n[P^2]_{ij}=sum_{j=1}^nsum_{k=1}^np_{ik}p_{kj}=sum_{k=1}^np_{ik}sum_{j=1}^np_{kj}=sum_{k=1}^np_{ik}cdot 1=1.$$
Then the result for $P^m$ follows by induction.
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
add a comment |
up vote
0
down vote
A generalization: if $P$ and $Q$ are two such matrices, the rows of $PQ$ also sum to $1$. For with
$P = [p_{ij}] tag{1}$
and
$Q = [q_{ij}] tag{2}$
with
$sum_j p_{ij} = sum_j q_{ij} = 1, tag{3}$
then
$sum_k (PQ)_{ik} = sum_k sum_j p_{ij}q_{jk} = sum_j sum_k p_{ij} q_{jk} = sum_j p_{ij} sum_k q_{jk} = sum_j p_{ij} = 1. tag{4}$
From this it follows, via a simple induction, that if $P_l$, $1 le l le m$, are $m$ such matrices, then
$sum_j (prod_1^m P_l)_{ij} = 1, tag{5}$
for clearly if (5) holds for some $m$, then taking $P = prod_1^m P_l$ and $Q = P_{m + 1}$ we may apply (4) to conclude that the row sums of $prod_1^{m + 1} P_l$ are also $1$; (4) forms a base for the induction. Now taking $P_l = P$, $1 le l le m$ for any $m$ yields the specific result requested in the text of the question:
$sum_j (P^m)_{ij} = 1. tag{6}$
QED.
Note: While were on the theme of generalization, it should be observed that the entries of the $P_l$ may be taken $Bbb Z$, $Bbb Q$, $Bbb R$, or $Bbb C$, and the result will bind. Leaping further, the assertion proved here holds as long as the entries of the $P_l$ are taken from any unital ring $R$, commutative or not, from $Bbb Z_2$ to $B(X)$, the set of bounded operators on a Banach space $X$ and beyond, as long as ring $R$ has a unit, these matrices may be taking from any $M_n(R)$, and as long as the $P_l$ satisfy
$sum_j (P_l)_{ij} = 1_R, tag{7}$
so will their product. A result of quite broad scope, indeed! End of Note.
Hope this helps. Cheerio,
and as ever,
Fiat Lux!!!
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
add a comment |
up vote
0
down vote
Since every row of $P$ sums to $1$, $v=[1,1,...,1]^T$ is an eigenvector of $P$ with eigenvalue $1$. Thus $Pv=vimplies P^2v=P(Pv)=Pv=v implies P^mv=v$, thus every row of $P^m$ sums to $1$ .
answered Dec 5 at 9:58
Rhaldryn
160213
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The row sums of any power of $P$ are always 1. To see this, write $[P]_{ij}:=p_{ij}$. Then
$$[P^2]_{ij}=sum_{k}^np_{ik}p_{kj}$$
and
$$sum_{j=1}^n[P^2]_{ij}=sum_{j=1}^nsum_{k=1}^np_{ik}p_{kj}=sum_{k=1}^np_{ik}sum_{j=1}^np_{kj}=sum_{k=1}^np_{ik}cdot 1=1.$$
Then the result for $P^m$ follows by induction.
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
add a comment |
up vote
3
down vote
accepted
The row sums of any power of $P$ are always 1. To see this, write $[P]_{ij}:=p_{ij}$. Then
$$[P^2]_{ij}=sum_{k}^np_{ik}p_{kj}$$
and
$$sum_{j=1}^n[P^2]_{ij}=sum_{j=1}^nsum_{k=1}^np_{ik}p_{kj}=sum_{k=1}^np_{ik}sum_{j=1}^np_{kj}=sum_{k=1}^np_{ik}cdot 1=1.$$
Then the result for $P^m$ follows by induction.
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The row sums of any power of $P$ are always 1. To see this, write $[P]_{ij}:=p_{ij}$. Then
$$[P^2]_{ij}=sum_{k}^np_{ik}p_{kj}$$
and
$$sum_{j=1}^n[P^2]_{ij}=sum_{j=1}^nsum_{k=1}^np_{ik}p_{kj}=sum_{k=1}^np_{ik}sum_{j=1}^np_{kj}=sum_{k=1}^np_{ik}cdot 1=1.$$
Then the result for $P^m$ follows by induction.
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
The row sums of any power of $P$ are always 1. To see this, write $[P]_{ij}:=p_{ij}$. Then
$$[P^2]_{ij}=sum_{k}^np_{ik}p_{kj}$$
and
$$sum_{j=1}^n[P^2]_{ij}=sum_{j=1}^nsum_{k=1}^np_{ik}p_{kj}=sum_{k=1}^np_{ik}sum_{j=1}^np_{kj}=sum_{k=1}^np_{ik}cdot 1=1.$$
Then the result for $P^m$ follows by induction.
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
edited Oct 20 '14 at 3:43
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
answered Oct 20 '14 at 3:24
Alex R.
24.7k12352
24.7k12352
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
add a comment |
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
@shooting-squirrel: The row sums are equal to 1, meaning that for any $i$, $sum_{j=1}^np_{ij}=1$. What is unclear about this?
– Alex R.
Oct 20 '14 at 3:42
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
Sorry, my mistake.
– shooting-squirrel
Oct 20 '14 at 3:49
add a comment |
up vote
0
down vote
A generalization: if $P$ and $Q$ are two such matrices, the rows of $PQ$ also sum to $1$. For with
$P = [p_{ij}] tag{1}$
and
$Q = [q_{ij}] tag{2}$
with
$sum_j p_{ij} = sum_j q_{ij} = 1, tag{3}$
then
$sum_k (PQ)_{ik} = sum_k sum_j p_{ij}q_{jk} = sum_j sum_k p_{ij} q_{jk} = sum_j p_{ij} sum_k q_{jk} = sum_j p_{ij} = 1. tag{4}$
From this it follows, via a simple induction, that if $P_l$, $1 le l le m$, are $m$ such matrices, then
$sum_j (prod_1^m P_l)_{ij} = 1, tag{5}$
for clearly if (5) holds for some $m$, then taking $P = prod_1^m P_l$ and $Q = P_{m + 1}$ we may apply (4) to conclude that the row sums of $prod_1^{m + 1} P_l$ are also $1$; (4) forms a base for the induction. Now taking $P_l = P$, $1 le l le m$ for any $m$ yields the specific result requested in the text of the question:
$sum_j (P^m)_{ij} = 1. tag{6}$
QED.
Note: While were on the theme of generalization, it should be observed that the entries of the $P_l$ may be taken $Bbb Z$, $Bbb Q$, $Bbb R$, or $Bbb C$, and the result will bind. Leaping further, the assertion proved here holds as long as the entries of the $P_l$ are taken from any unital ring $R$, commutative or not, from $Bbb Z_2$ to $B(X)$, the set of bounded operators on a Banach space $X$ and beyond, as long as ring $R$ has a unit, these matrices may be taking from any $M_n(R)$, and as long as the $P_l$ satisfy
$sum_j (P_l)_{ij} = 1_R, tag{7}$
so will their product. A result of quite broad scope, indeed! End of Note.
Hope this helps. Cheerio,
and as ever,
Fiat Lux!!!
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
add a comment |
up vote
0
down vote
A generalization: if $P$ and $Q$ are two such matrices, the rows of $PQ$ also sum to $1$. For with
$P = [p_{ij}] tag{1}$
and
$Q = [q_{ij}] tag{2}$
with
$sum_j p_{ij} = sum_j q_{ij} = 1, tag{3}$
then
$sum_k (PQ)_{ik} = sum_k sum_j p_{ij}q_{jk} = sum_j sum_k p_{ij} q_{jk} = sum_j p_{ij} sum_k q_{jk} = sum_j p_{ij} = 1. tag{4}$
From this it follows, via a simple induction, that if $P_l$, $1 le l le m$, are $m$ such matrices, then
$sum_j (prod_1^m P_l)_{ij} = 1, tag{5}$
for clearly if (5) holds for some $m$, then taking $P = prod_1^m P_l$ and $Q = P_{m + 1}$ we may apply (4) to conclude that the row sums of $prod_1^{m + 1} P_l$ are also $1$; (4) forms a base for the induction. Now taking $P_l = P$, $1 le l le m$ for any $m$ yields the specific result requested in the text of the question:
$sum_j (P^m)_{ij} = 1. tag{6}$
QED.
Note: While were on the theme of generalization, it should be observed that the entries of the $P_l$ may be taken $Bbb Z$, $Bbb Q$, $Bbb R$, or $Bbb C$, and the result will bind. Leaping further, the assertion proved here holds as long as the entries of the $P_l$ are taken from any unital ring $R$, commutative or not, from $Bbb Z_2$ to $B(X)$, the set of bounded operators on a Banach space $X$ and beyond, as long as ring $R$ has a unit, these matrices may be taking from any $M_n(R)$, and as long as the $P_l$ satisfy
$sum_j (P_l)_{ij} = 1_R, tag{7}$
so will their product. A result of quite broad scope, indeed! End of Note.
Hope this helps. Cheerio,
and as ever,
Fiat Lux!!!
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
add a comment |
up vote
0
down vote
up vote
0
down vote
A generalization: if $P$ and $Q$ are two such matrices, the rows of $PQ$ also sum to $1$. For with
$P = [p_{ij}] tag{1}$
and
$Q = [q_{ij}] tag{2}$
with
$sum_j p_{ij} = sum_j q_{ij} = 1, tag{3}$
then
$sum_k (PQ)_{ik} = sum_k sum_j p_{ij}q_{jk} = sum_j sum_k p_{ij} q_{jk} = sum_j p_{ij} sum_k q_{jk} = sum_j p_{ij} = 1. tag{4}$
From this it follows, via a simple induction, that if $P_l$, $1 le l le m$, are $m$ such matrices, then
$sum_j (prod_1^m P_l)_{ij} = 1, tag{5}$
for clearly if (5) holds for some $m$, then taking $P = prod_1^m P_l$ and $Q = P_{m + 1}$ we may apply (4) to conclude that the row sums of $prod_1^{m + 1} P_l$ are also $1$; (4) forms a base for the induction. Now taking $P_l = P$, $1 le l le m$ for any $m$ yields the specific result requested in the text of the question:
$sum_j (P^m)_{ij} = 1. tag{6}$
QED.
Note: While were on the theme of generalization, it should be observed that the entries of the $P_l$ may be taken $Bbb Z$, $Bbb Q$, $Bbb R$, or $Bbb C$, and the result will bind. Leaping further, the assertion proved here holds as long as the entries of the $P_l$ are taken from any unital ring $R$, commutative or not, from $Bbb Z_2$ to $B(X)$, the set of bounded operators on a Banach space $X$ and beyond, as long as ring $R$ has a unit, these matrices may be taking from any $M_n(R)$, and as long as the $P_l$ satisfy
$sum_j (P_l)_{ij} = 1_R, tag{7}$
so will their product. A result of quite broad scope, indeed! End of Note.
Hope this helps. Cheerio,
and as ever,
Fiat Lux!!!
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
A generalization: if $P$ and $Q$ are two such matrices, the rows of $PQ$ also sum to $1$. For with
$P = [p_{ij}] tag{1}$
and
$Q = [q_{ij}] tag{2}$
with
$sum_j p_{ij} = sum_j q_{ij} = 1, tag{3}$
then
$sum_k (PQ)_{ik} = sum_k sum_j p_{ij}q_{jk} = sum_j sum_k p_{ij} q_{jk} = sum_j p_{ij} sum_k q_{jk} = sum_j p_{ij} = 1. tag{4}$
From this it follows, via a simple induction, that if $P_l$, $1 le l le m$, are $m$ such matrices, then
$sum_j (prod_1^m P_l)_{ij} = 1, tag{5}$
for clearly if (5) holds for some $m$, then taking $P = prod_1^m P_l$ and $Q = P_{m + 1}$ we may apply (4) to conclude that the row sums of $prod_1^{m + 1} P_l$ are also $1$; (4) forms a base for the induction. Now taking $P_l = P$, $1 le l le m$ for any $m$ yields the specific result requested in the text of the question:
$sum_j (P^m)_{ij} = 1. tag{6}$
QED.
Note: While were on the theme of generalization, it should be observed that the entries of the $P_l$ may be taken $Bbb Z$, $Bbb Q$, $Bbb R$, or $Bbb C$, and the result will bind. Leaping further, the assertion proved here holds as long as the entries of the $P_l$ are taken from any unital ring $R$, commutative or not, from $Bbb Z_2$ to $B(X)$, the set of bounded operators on a Banach space $X$ and beyond, as long as ring $R$ has a unit, these matrices may be taking from any $M_n(R)$, and as long as the $P_l$ satisfy
$sum_j (P_l)_{ij} = 1_R, tag{7}$
so will their product. A result of quite broad scope, indeed! End of Note.
Hope this helps. Cheerio,
and as ever,
Fiat Lux!!!
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
answered Oct 20 '14 at 6:13
Robert Lewis
43k22863
43k22863
add a comment |
add a comment |
up vote
0
down vote
Since every row of $P$ sums to $1$, $v=[1,1,...,1]^T$ is an eigenvector of $P$ with eigenvalue $1$. Thus $Pv=vimplies P^2v=P(Pv)=Pv=v implies P^mv=v$, thus every row of $P^m$ sums to $1$ .
answered Dec 5 at 9:58
Rhaldryn
160213
add a comment |
up vote
0
down vote
Since every row of $P$ sums to $1$, $v=[1,1,...,1]^T$ is an eigenvector of $P$ with eigenvalue $1$. Thus $Pv=vimplies P^2v=P(Pv)=Pv=v implies P^mv=v$, thus every row of $P^m$ sums to $1$ .
answered Dec 5 at 9:58
Rhaldryn
160213
add a comment |
up vote
0
down vote
up vote
0
down vote
Since every row of $P$ sums to $1$, $v=[1,1,...,1]^T$ is an eigenvector of $P$ with eigenvalue $1$. Thus $Pv=vimplies P^2v=P(Pv)=Pv=v implies P^mv=v$, thus every row of $P^m$ sums to $1$ .
answered Dec 5 at 9:58
Rhaldryn
160213
Since every row of $P$ sums to $1$, $v=[1,1,...,1]^T$ is an eigenvector of $P$ with eigenvalue $1$. Thus $Pv=vimplies P^2v=P(Pv)=Pv=v implies P^mv=v$, thus every row of $P^m$ sums to $1$ .
answered Dec 5 at 9:58
Rhaldryn
160213
answered Dec 5 at 9:58
Rhaldryn
160213
answered Dec 5 at 9:58
Rhaldryn
160213
answered Dec 5 at 9:58
Rhaldryn
160213
160213
add a comment |
add a comment |
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Do you mean that the sum of each row is equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:21
What kind of entries does P have?i.e. real, complex, integers?
– shooting-squirrel
Oct 20 '14 at 3:21
entries are real and the row sums=1 means each row adds to 1
– Learnmore
Oct 20 '14 at 3:23
So all the values on a row, added up, equal to $1$?
– shooting-squirrel
Oct 20 '14 at 3:24
Yes all the values on a row, added up, equal to 1
– Learnmore
Oct 20 '14 at 3:43