Creating a steady state vector
$begingroup$
I'm confused on where the intuition came from to divide $w$ by the sum of its entries to find $q$. I don't really see the relation from the sum of its entries with "every solution being a multiple of the solution $w$".
linear-algebra probability vector-spaces steady-state
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
$endgroup$
add a comment |
$begingroup$
I'm confused on where the intuition came from to divide $w$ by the sum of its entries to find $q$. I don't really see the relation from the sum of its entries with "every solution being a multiple of the solution $w$".
linear-algebra probability vector-spaces steady-state
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
$endgroup$
2
$begingroup$
If we want $q$ to represent a probability vector, then we want every entry of $q$ to be in the interval $[0,1]$ and to have all of the entries add up to one. In order to make it add up to one, this is easily accomplished by dividing by the sum of the entries since $frac{w_1}{w_1+w_2+dots+w_n}+frac{w_2}{w_1+w_2+dots+w_n}+dots+frac{w_n}{w_1+w_2+dots+w_n}=frac{w_1+w_2+dots+w_n}{w_1+w_2+dots+w_n}=1$
$endgroup$
– JMoravitz
Mar 14 '17 at 6:04
$begingroup$
As for every solution being a multiple of $w$ (or a multiple of $q$ for that matter), that is a result of the eigenspace corresponding to the eigenvector of $1$ for any normal stochastic matrix will be one-dimensional. Any one-dimensional space you have all vectors in the space (in this case, our space of steadystate vectors) will be multiples of one another (except for being a multiple of the zero vector). This is unrelated to the sum of the entries.
$endgroup$
– JMoravitz
Mar 14 '17 at 6:09
add a comment |
$begingroup$
I'm confused on where the intuition came from to divide $w$ by the sum of its entries to find $q$. I don't really see the relation from the sum of its entries with "every solution being a multiple of the solution $w$".
linear-algebra probability vector-spaces steady-state
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
$endgroup$
I'm confused on where the intuition came from to divide $w$ by the sum of its entries to find $q$. I don't really see the relation from the sum of its entries with "every solution being a multiple of the solution $w$".
linear-algebra probability vector-spaces steady-state
linear-algebra probability vector-spaces steady-state
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
asked Mar 14 '17 at 5:53
stumpedstumped
5731824
5731824
2
$begingroup$
If we want $q$ to represent a probability vector, then we want every entry of $q$ to be in the interval $[0,1]$ and to have all of the entries add up to one. In order to make it add up to one, this is easily accomplished by dividing by the sum of the entries since $frac{w_1}{w_1+w_2+dots+w_n}+frac{w_2}{w_1+w_2+dots+w_n}+dots+frac{w_n}{w_1+w_2+dots+w_n}=frac{w_1+w_2+dots+w_n}{w_1+w_2+dots+w_n}=1$
$endgroup$
– JMoravitz
Mar 14 '17 at 6:04
$begingroup$
As for every solution being a multiple of $w$ (or a multiple of $q$ for that matter), that is a result of the eigenspace corresponding to the eigenvector of $1$ for any normal stochastic matrix will be one-dimensional. Any one-dimensional space you have all vectors in the space (in this case, our space of steadystate vectors) will be multiples of one another (except for being a multiple of the zero vector). This is unrelated to the sum of the entries.
$endgroup$
– JMoravitz
Mar 14 '17 at 6:09
add a comment |
2
$begingroup$
If we want $q$ to represent a probability vector, then we want every entry of $q$ to be in the interval $[0,1]$ and to have all of the entries add up to one. In order to make it add up to one, this is easily accomplished by dividing by the sum of the entries since $frac{w_1}{w_1+w_2+dots+w_n}+frac{w_2}{w_1+w_2+dots+w_n}+dots+frac{w_n}{w_1+w_2+dots+w_n}=frac{w_1+w_2+dots+w_n}{w_1+w_2+dots+w_n}=1$
$endgroup$
– JMoravitz
Mar 14 '17 at 6:04
$begingroup$
As for every solution being a multiple of $w$ (or a multiple of $q$ for that matter), that is a result of the eigenspace corresponding to the eigenvector of $1$ for any normal stochastic matrix will be one-dimensional. Any one-dimensional space you have all vectors in the space (in this case, our space of steadystate vectors) will be multiples of one another (except for being a multiple of the zero vector). This is unrelated to the sum of the entries.
$endgroup$
– JMoravitz
Mar 14 '17 at 6:09
2
2
$begingroup$
If we want $q$ to represent a probability vector, then we want every entry of $q$ to be in the interval $[0,1]$ and to have all of the entries add up to one. In order to make it add up to one, this is easily accomplished by dividing by the sum of the entries since $frac{w_1}{w_1+w_2+dots+w_n}+frac{w_2}{w_1+w_2+dots+w_n}+dots+frac{w_n}{w_1+w_2+dots+w_n}=frac{w_1+w_2+dots+w_n}{w_1+w_2+dots+w_n}=1$
$endgroup$
– JMoravitz
Mar 14 '17 at 6:04
$begingroup$
If we want $q$ to represent a probability vector, then we want every entry of $q$ to be in the interval $[0,1]$ and to have all of the entries add up to one. In order to make it add up to one, this is easily accomplished by dividing by the sum of the entries since $frac{w_1}{w_1+w_2+dots+w_n}+frac{w_2}{w_1+w_2+dots+w_n}+dots+frac{w_n}{w_1+w_2+dots+w_n}=frac{w_1+w_2+dots+w_n}{w_1+w_2+dots+w_n}=1$
$endgroup$
– JMoravitz
Mar 14 '17 at 6:04
$begingroup$
As for every solution being a multiple of $w$ (or a multiple of $q$ for that matter), that is a result of the eigenspace corresponding to the eigenvector of $1$ for any normal stochastic matrix will be one-dimensional. Any one-dimensional space you have all vectors in the space (in this case, our space of steadystate vectors) will be multiples of one another (except for being a multiple of the zero vector). This is unrelated to the sum of the entries.
$endgroup$
– JMoravitz
Mar 14 '17 at 6:09
$begingroup$
As for every solution being a multiple of $w$ (or a multiple of $q$ for that matter), that is a result of the eigenspace corresponding to the eigenvector of $1$ for any normal stochastic matrix will be one-dimensional. Any one-dimensional space you have all vectors in the space (in this case, our space of steadystate vectors) will be multiples of one another (except for being a multiple of the zero vector). This is unrelated to the sum of the entries.
$endgroup$
– JMoravitz
Mar 14 '17 at 6:09
add a comment |
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In comments the user JMoravitz gives an answer. The solution to the question says that he wants to find the probability set of all solutions. So he finds a vector that solves the equation: $vec{w} = (3,4)$ and then you normalize it. So the entries of $vec{q}$.
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
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$begingroup$
In comments the user JMoravitz gives an answer. The solution to the question says that he wants to find the probability set of all solutions. So he finds a vector that solves the equation: $vec{w} = (3,4)$ and then you normalize it. So the entries of $vec{q}$.
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
$endgroup$
add a comment |
$begingroup$
In comments the user JMoravitz gives an answer. The solution to the question says that he wants to find the probability set of all solutions. So he finds a vector that solves the equation: $vec{w} = (3,4)$ and then you normalize it. So the entries of $vec{q}$.
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
$endgroup$
add a comment |
$begingroup$
In comments the user JMoravitz gives an answer. The solution to the question says that he wants to find the probability set of all solutions. So he finds a vector that solves the equation: $vec{w} = (3,4)$ and then you normalize it. So the entries of $vec{q}$.
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
$endgroup$
In comments the user JMoravitz gives an answer. The solution to the question says that he wants to find the probability set of all solutions. So he finds a vector that solves the equation: $vec{w} = (3,4)$ and then you normalize it. So the entries of $vec{q}$.
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
answered Mar 29 '17 at 4:05
Rafael WagnerRafael Wagner
1,8332923
1,8332923
add a comment |
add a comment |
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$begingroup$
If we want $q$ to represent a probability vector, then we want every entry of $q$ to be in the interval $[0,1]$ and to have all of the entries add up to one. In order to make it add up to one, this is easily accomplished by dividing by the sum of the entries since $frac{w_1}{w_1+w_2+dots+w_n}+frac{w_2}{w_1+w_2+dots+w_n}+dots+frac{w_n}{w_1+w_2+dots+w_n}=frac{w_1+w_2+dots+w_n}{w_1+w_2+dots+w_n}=1$
$endgroup$
– JMoravitz
Mar 14 '17 at 6:04
$begingroup$
As for every solution being a multiple of $w$ (or a multiple of $q$ for that matter), that is a result of the eigenspace corresponding to the eigenvector of $1$ for any normal stochastic matrix will be one-dimensional. Any one-dimensional space you have all vectors in the space (in this case, our space of steadystate vectors) will be multiples of one another (except for being a multiple of the zero vector). This is unrelated to the sum of the entries.
$endgroup$
– JMoravitz
Mar 14 '17 at 6:09