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Vector sequences


Thread replies: 12
Thread images: 3

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How do you define "increasing" or "decreasing" for a sequence of vectors? Do we use the standard norm [math]|| \cdot ||: V \rightarrow \mathbb{R} [/math] or what?
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>>7745062
Considering you just made up those two meaningless terms you can apply whatever definition you want.
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is length usually drawn with 2 lines or 1?
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>>7745065
W-What did I make up?
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>>7745068
We do it with 2 lines breh
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>>7745062
You can say the sequence of norms is increasing. No need to bother with imprecise and clumsy extensions of familiar definitions to absurd contexts.
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>>7745071
But breh I need to prove the following sequence is converging and I want to do it with the monotone convergence theorem:

[eqn] \lim_{k\to\infty} \mathbf{v}P^{k} [/eqn]

I already know the sequence is bounded.
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>>7745077
What is P here? In any case I don't think there's any analogue of the theorem you're thinking of outside of R, so you'll need to use something else.
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>>7745078
P is a matrix representing a Markov chain. However P isn't diagonalizable. If we know the sequence converges it's easy. We just call [math] \lim_{k\to\infty} \mathbf{v}P^{k} = \mathbf{x} [/math] and then say:

[eqn] \lim_{k\to\infty} \mathbf{v}P^{k} = \lim_{k\to\infty} \mathbf{v}P^{k+1} \\

\lim_{k\to\infty} \mathbf{v}P^{k+1} = P \dot \lim_{k\to\infty} \mathbf{v}P^{k} \\

\mathbf{x} = \mathbf{x}P [/eqn]

Where [math]\mathbf{x}[/math] is the equilibrium vector but first we have to prove it converges it's driving me freaking nuts.
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>>7745087
I'm probably missing something but this seems obvious if you use a Jordan basis for P.
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>>7745095
>a Jordan basis for P
I haven't learned this yet m8
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>>7745098
Never mind, I realized that the theorem I was thinking of only works for complex vector spaces. I'm pretty sure this isn't true in general for operators on a real vector space so you'll have to use things about Markov chains specifically, which I don't know much about.
Thread replies: 12
Thread images: 3
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