User FuzzyPixelz | chat.stackexchange.com

FuzzyPixelz

Apr 26, 2022 13:31

Is there a special name for First-Order Logic with all terms restricted to variables (i.e no function terms)? Is it simpler to decide satisfiability in this case?

FuzzyPixelz

Jan 8, 2022 13:57

For context, I was reading a Type Theory paper and the author used a join-semilattice only to make the join operator on it partial, which to my mind is just a partial order "with extra steps".

FuzzyPixelz

Jan 8, 2022 13:42

Lattice (order)

A lattice is an abstract structure studied in the mathematical subdisciplines of order theory and abstract algebra. It consists of a partially ordered set in which every pair of elements has a unique supremum (also called a least upper bound or join) and a unique infimum (also called a greatest lower bound or meet). An example is given by the power set of a set, partially ordered by inclusion, for which the supremum is the union and the infimum is the intersection. Another example is given by the natural numbers, partially ordered by divisibility, for which the supremum is the least common multiple...

FuzzyPixelz

Jan 8, 2022 13:40

If any two elements of a join-semilattice are comparable by definition, then the set is not a poset anymore, so why define a join-semilattice in terms of a poset?

FuzzyPixelz

Apr 6, 2020 12:33

Yeah module theory, never touched it

FuzzyPixelz

Apr 6, 2020 12:31

You sound surprised

FuzzyPixelz

Apr 6, 2020 12:30

Sounds right?

FuzzyPixelz

Apr 6, 2020 12:30

So $\ker(A-\lambda_i I)^{l}=\ker(A-\lambda_i I)^{m_i}$ if and only if $l \ge m_i$ where $m_i$ is the multiplicity of $\lambda_i$ in the minimal polynomial $\mu_A$.

FuzzyPixelz

Apr 6, 2020 12:30

So by Primary Decomposition we can write $V=\ker(A-\lambda_1 I)^{\alpha_1} \oplus \cdots \oplus \ker(A-\lambda_r I)^{\alpha_r}$. So if we consider $A_i$ the restriction of $A$ on $\ker(A-\lambda_i I)^{\alpha_i}$, its characteristic polynomial is $\chi_{A_i}(t)=(t-\lambda_i)^{n_i}$ where $n_i=\dim \ker(A-\lambda_i I)^{\alpha_i}$ so the characteristic polynomial of $A$ is $\chi_A=(t-\lambda_1)^{n_1}\cdots (t-\lambda_r)^{n_r}$, hence $n_i$ is multiplicity of $\lambda_i$ in $\chi_A$.

FuzzyPixelz

Apr 6, 2020 12:30

Just checking some things. If a polynomial $f(t) \in \mathbb C[t]$ annihilates $A:\mathbb C^n \to \mathbb C^n$, then every eigen value of $A$ is a root of $f(t)$ so we can factorize $f(t)=(t-\lambda_1)^{\alpha_1}\cdots (t-\lambda_r)^{\alpha_r}$ where the $\lambda_i$ are the unique eigenvalues.

FuzzyPixelz

Apr 6, 2020 06:46

Can you point me to a proof of the existence of JNF that explains in detail how you would go about finding the form? :P

FuzzyPixelz

Apr 5, 2020 22:06

Most proofs I've seen use weird arguments for existence and then handwave their way through examples

FuzzyPixelz

Apr 5, 2020 22:04

I want a rigorous proof

FuzzyPixelz

Apr 5, 2020 22:04

I know the steps at least

FuzzyPixelz

Apr 5, 2020 22:03

I still don't understand why we can always find Jordan Chains, but yes

FuzzyPixelz

Apr 5, 2020 21:55

Which must be a proper basis of $V$, does this tell you enough @TedShifrin ?

FuzzyPixelz

Apr 5, 2020 21:54

For induction it considers $\text{Im}A$

FuzzyPixelz

Apr 5, 2020 21:53

Then it follows by induction on the dimension of $V$, proving that there is always a basis of eigenvectors $\{v_1, \dots, v_n\}$ s.t. $Av_i$ is either $v_{i+1}$ or $0$

FuzzyPixelz

Apr 5, 2020 21:51

First, it reduces to the case where $A$ is nilpotent (Primary Decomposition Th. and adding $\lambda Id$)

FuzzyPixelz

Apr 5, 2020 21:33

... Can someone point me to a proof of the existence of JNF that actually also explains how to obtain it :)

FuzzyPixelz

Apr 5, 2020 09:37

That's fair, I think book doesn't look too bad :P

FuzzyPixelz

Apr 5, 2020 09:30

@AlessandroCodenotti I hope

FuzzyPixelz

Apr 5, 2020 09:30

Alrighty, thanks.

FuzzyPixelz

Apr 5, 2020 09:29

Something relying on something else a finite dimensional vector space over a specific field is already too much for me to telerate

FuzzyPixelz

Apr 5, 2020 09:26

@LeakyNun do you recommend any concise and detailed book?

FuzzyPixelz

Apr 5, 2020 09:25

All I see you doing all day everyday is algebra @BalarkaSen

FuzzyPixelz

Apr 5, 2020 09:24

Haha

FuzzyPixelz

Apr 5, 2020 09:19

The fact that I don't know any of that makes me feel better actually

FuzzyPixelz

Apr 5, 2020 09:17

$A$ is a linear map on a finite dimensional vector space $V$ over the complex numbers.

FuzzyPixelz

Apr 5, 2020 09:16

I'm not much of an abstract algebra nor arithmetic person so please tell me why would you or anyone else think like this?

FuzzyPixelz

Apr 5, 2020 09:16

If $v \neq 0$ is $(A-\alpha I)$-cyclic, with period $r$, then $\{ v, (A-\alpha I)v, \dots, (A-\alpha I)^{r-1}v\}$ is linearly independent. My proof consists of repeatedly composing by $A^{r-1}$, $A^{r}$... My textbook however writes the linear dependence relation as $f(A-\alpha I)v=0$ where $f$ is a polynomial, and straight out talks about the greatest common divisor of $f(t)$ and $g(t)=t^r$.

FuzzyPixelz

Mar 7, 2020 20:26

You mean the zeros I added? I only did it for the integral with $x$ not $x_0$

FuzzyPixelz

Mar 7, 2020 20:24

Another error?

FuzzyPixelz

Mar 7, 2020 20:23

Thanks @TedShifrin you're awesome.

FuzzyPixelz

Mar 7, 2020 20:22

Is what I did

FuzzyPixelz

Mar 7, 2020 20:22

$$\int_y^z=\int_y^{y_0} + \int_{y_0}^{z_0}+\int_{z_0}^{z}$$

FuzzyPixelz

Mar 7, 2020 20:18

(Hopefully)

FuzzyPixelz

Mar 7, 2020 20:17

Not really Lipschitz continuous this time, but it's continuous

FuzzyPixelz

Mar 7, 2020 20:14

So the LHS becomes less than or equal to $$\max(M, k|y_0-z_0|) \| (x,y,z)- (x_0,y_0,z_0)\|$$

FuzzyPixelz

Mar 7, 2020 20:14

$$\left | \frac{\partial \phi}{\partial x}(x,y,z)-\frac{\partial \phi}{\partial x}(x_0,y_0,z_0) \right |=\left |\int_{z_0}^z \frac{\partial f}{\partial x}(x,t)dt - \int_{y_0}^{y} \frac{\partial f}{\partial x}(x,t)dt +\int_{y_0}^{z_0} \left (\frac{\partial f}{\partial x}(x,t)-\frac{\partial f}{\partial x}(x_0,t) \right )dt \right |$$ Since $\partial f/ \partial x$ is uniformly continuous we can write $| \frac{\partial f}{\partial x}(x,t)-\frac{\partial f}{\partial x}(x_0,t)| \le k |x-x_0|$

FuzzyPixelz

Mar 7, 2020 19:55

I'll do it now

FuzzyPixelz

Mar 7, 2020 19:54

That would add $M|x-x_0|$ on the RHS

FuzzyPixelz

Mar 7, 2020 19:53

I would have to use the uniform continuity of $\frac{\partial f}{\partial x}$ like you said @TedShifrin

FuzzyPixelz

Mar 7, 2020 19:52

I see.. it's definitely wrong

FuzzyPixelz

Mar 7, 2020 19:49

Oh, I should've replaced $x$ with $x_0$ in the $\frac{\partial f}{\partial x}(x,t)$ inside the second integral, but I only used the fact that $\frac{\partial f}{\partial x}$ is bounded afterwards so it should be fine?

FuzzyPixelz

Mar 7, 2020 19:46

I'm sorry, what is my mistake exactly

FuzzyPixelz

Mar 7, 2020 19:45

I wrote an answer down..

FuzzyPixelz

Mar 7, 2020 19:45

Wait, are we talking about the first or the second post?

FuzzyPixelz

Mar 7, 2020 19:41

I used the max norm?

FuzzyPixelz

Mar 7, 2020 19:38

I didn't get stuck, I'm just looking to verify my proof

FuzzyPixelz

Mathematics

$Mathematics$ Mathematics