Stochastic Matrices

>>7699140
Start with x. Multiply by P. Multiply the result by P. Repeat until you have done k multiplications.

>>7699168
Come on now

>>7699140
Is P nilpotent?

>>7699187
You tell me. Is that possible with stochastic matrices?

>>7699195
Fuck if I know, I thought you were teaching us.

>>7699140
>>stackoverflow

>>7699222
I don't think it's possible for a stochastic matrix to be nilpotent. Consider the following:

For every vector [math]\textbf{x}[/math] in [math]\mathbb{R}^{n}[/math] with positive entries, we can multiply it by [math]\frac{1]{||\textbf{x}||}[/math] to get a stochastic vector. Call this scalar [math]c[/math].

So now we have:

[math]cP^{k}\textbf{x}[/math]

And when [math]P[/math] is a stochastic matrix and [math]\textbf{x}[/math] a stochastic vector, the result will always be a stochastic vector. Therefore, [math]P^{k}[/math] can't be zero. for any [math]k \in \mathbb{N}[/math].

But this only takes into account vectors with strictly positive entries.

>>7699235
This seems right, P^k is never zero. P^k * 1 = 1 for any k, where 1 is the vector of ones. This is because the row sum of P is 1.

>>7699251
Bump I'm drunk now so I made up a better proof, check this out:

For any vector [math]\textbf{x}[/math] in [math]\mathbb{R}^{n}[/math] except zero, we can write it as:

[eqn] \textbf{x} = \textbf{x_{1}}+\textbf{x_{2}}[/eqn]

Where [math]\textbf{x_1}[/math] has strict positive entries and [math]\textbf{x_2}[/math] (the proof is trivial and left to the reader). Now we can rewrite those vectors as:

[eqn] \textbf{x} = ||\textbf{x_1}|| \frac{1}{||\textbf{x_1}||} \textbf{x_1} -||\textbf{x_2}|| \frac{1}{||\textbf{x_2}||} \textbf{x_2} [/eqn]

As you can see [eqn]\frac{1}{||\textbf{x_i}||} \textbf{x_i}[/eqn] is a stochastic vector. So then we have:

[eqn] P^{k} \textbf{x} = ||\textbf{x_1}|| P^{k} \frac{1}{||\textbf{x_1}||} \textbf{x_1} -||\textbf{x_2}|| P^{k} \frac{1}{||\textbf{x_2}||} \textbf{x_2}

P^{k} \textbf{x} = || \textbf{x_1}|| \textbf{v_1} -|| \textbf{x_2} || \textbf{v_2}
[/eqn]

Where [math]v_1[/math] and [math]v_2[/math] are stochastic vectors.

So we see [math]P^{k}\textbf{x}[/math] can't be zero except for when:

[eqn] ||\textbf{x_1}|| \textbf{v_1} = || \textbf{x_2} || \textbf{v_2} [/eqn]

But we assumed [math]\textbf{x} \neq \textbf{0}[/math], so this never happens, so

[eqn] P^{k} \textbf{x} \neq \textbf{0} [/eqn]

For any [math] k \in \mathbb{N}[/math].

So any stochastic matrix [math]P[/math] isn't nilpotent.

How is my rigorous proof?

>>7699658
>Where x_1 has strict positive entries and x_2 (the proof is trivial and left to the reader)
And x_2 strict negative entries.

>>7699658
Did I fucked up the LaTeX again?

Shouldn't have used the subscript on the textbf.

last bump before i go to sleep

>>7699658
Let x_1=<1,1> in R^2 with the standard norm. Then x_1 has strictly positive entries but (1/||x_1||)x_1=<1/sqrt(2), 1/sqrt(2)> which is not a stochastic vector.

Are you assuming a particular norm?

>>7701356
Damn, you're right. I was thinking about unit vectors. Anyway if we do

[eqn] c = \sum_i \textbf{x}_i [/eqn]

Then [math]\frac{1}{c} \textbf{x}[/math] should be a stochastic vector.

>>7701558
Where the [math]\textbf{x}_i[/math] are the entries of the vector.

I doubt this is much easier than computing A^k x for some general matrix A. All we really know about P is that it has positive entries and it's row sum is 1. That gives us one eigenvector and value, but not much more.

If you want to compute A^k, the common method is repeated squaring.

P^k * 1 = 1 for all k tells us P^k is never zero.

>>7703729
Wew lad someone else bumping my thread?

This seems like an appropriate thread to ask another question; does anyone know of a measure-preserving transformation whose transfer operator is self-adjoint?