[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 103 KB, 750x600, 1428529584896.jpg [View same] [iqdb] [saucenao] [google]
7745062 No.7745062 [Reply] [Original]

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?

>> No.7745065

>>7745062
Considering you just made up those two meaningless terms you can apply whatever definition you want.

>> No.7745068

is length usually drawn with 2 lines or 1?

>> No.7745069
File: 3 KB, 125x119, sweating pepe.jpg [View same] [iqdb] [saucenao] [google]
7745069

>>7745065
W-What did I make up?

>> No.7745070

>>7745068
We do it with 2 lines breh

>> No.7745071

>>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.

>> No.7745077

>>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.

>> No.7745078

>>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.

>> No.7745087
File: 31 KB, 600x600, 1437783972309.jpg [View same] [iqdb] [saucenao] [google]
7745087

>>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.

>> No.7745095

>>7745087
I'm probably missing something but this seems obvious if you use a Jordan basis for P.

>> No.7745098

>>7745095
>a Jordan basis for P
I haven't learned this yet m8

>> No.7745127

>>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.