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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 $w$".

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stumped
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    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 $\frac{w_1}{w_1+w_2+\dots+w_n}+\frac{w_2}{w_1+w_2+\dots+w_n}+\dots+\frac{w_n}{w_1+w_2+\dots+w_n}=\frac{w_1+w_2+\dots+w_n}{w_1+w_2+\dots+w_n}=1$ – JMoravitz Mar 14 '17 at 06:04
  • 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. – JMoravitz Mar 14 '17 at 06:09

2 Answers2

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This question is quite old, but it's been viewed enough times where I think people are finding it to assist their learning and I feel the other answers aren't providing an explanation beneficial to help learn.

The first stage that you appeared fine with was actually solving out the equation to find a vector $x$ such that $$ Px = x. $$

The first question you had is why are all solutions scalar multiples. I'll show that if $k$ is a scalar then $kx$ is also a solution. Now scalar multiplication commutes and also associates, so we can see that $$ \begin{align} P(kx) = (Pk)x = k(Px) = kx \end{align} $$ since we know that $Px=x$. As such all scalar multiples are also solutions. As for why you cannot have a different vector linearly independent from $x$ as a solution, the null space of $(P-I)$ is one dimensional so if there was another linearly independent vector from $x$, the null space would be two dimensional which isn't possible.

Now to answer your second question of what is the intuition to divide by $w$?, remember that the steady state represents a probability of where you will be after a very very long time. Probabilities have to add up to $1$, so if we pick $$ q = \frac{x}{\sum_{i=1}^2 x_i} $$ we'll end up with a vector of values that has sum $1$.

Why in the book did they choose $w=[3,4]$ instead of something else before dividing? Because integers are pretty and they probably thought it would make the division seem more clear -- which it seems it wasn't haha

wjmccann
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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 $\vec{q}$.

R.W
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