Bob Pen

general recent conversations

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

Bob Pen

Oct 4, 2021 06:05

This is the book: book-wright-ma.github.io/Book-WM-20210422.pdf, page 89 - 91

Bob Pen

Oct 4, 2021 06:02

In the textbook I am reading right now, it says:

(Restricted Strong Convexity). The matrix A satisfies the restricted strong convexity (RSC) condition of order $k$, with parameters $\mu>0$, $\alpha \geq 1$, if for every $\mathrm{I}$ of size at most $k$ and for all nonzero $\boldsymbol{h}$ satisfying $\left\|\boldsymbol{h}_{\mathbf{| c}}\right\|_{I} \leq$ $\alpha\left\|\boldsymbol{h}_{I}\right\|_{1}$
$$
\|\boldsymbol{A} \boldsymbol{h}\|_{2}^{2} \geq \mu\|\boldsymbol{h}\|_{2}^{2}
$$

However this is said to satisfy the Nullspace property:

Bob Pen

Oct 4, 2021 02:54

$\left\|\left(\boldsymbol{X}^{T} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{T} \boldsymbol{v}\right\|_{2} \leq \frac{1}{\sigma_{\min }(\boldsymbol{X})}\|\boldsymbol{v}\|_{2}$ Does anyone have any hints as to how I can proceed in proving this inequality?

Bob Pen

Mar 4, 2021 14:51

Where there is an answer in the comments, but the answer says that there is a nice general explanation. So that got me wondering, does the minimum norm imply a better fit for a regression problem?

Bob Pen

Mar 4, 2021 14:50

I found this question where math.stackexchange.com/questions/3931144/…

Bob Pen

Mar 4, 2021 14:49

Is it then a worse regression ?

Bob Pen

Mar 4, 2021 14:48

But what if you minimize $||Ax-b||^2$ and you obtain an $x$ solution with a higher norm

Bob Pen

Mar 4, 2021 14:48

Let us say $Ax = b$ , where $A$ is rank-deficient

Bob Pen

Mar 4, 2021 14:47

Why does one want the solution to the minimum norm solution?

Bob Pen

Mar 4, 2021 14:46

Given a least squares regression problem

Bob Pen

Mar 4, 2021 14:46

Yeah

Bob Pen

Mar 4, 2021 14:45

Hey, I was just wondering. In Linear Algebra, why does one want minimum norm solution ?

Bob Pen

Oct 18, 2020 19:26

Thank you!

Bob Pen

Oct 18, 2020 19:26

Indeed!

Bob Pen

Oct 18, 2020 19:20

Any hints hehe?

Bob Pen

Oct 18, 2020 19:18

@skillpatrol With epsilon-delta?

Bob Pen

Oct 18, 2020 19:15

But I am still a little lost on how I can show that my limit goes to infinity

Bob Pen

Oct 18, 2020 19:14

Thanks for the explanation

Bob Pen

Oct 18, 2020 19:14

Indeed

Bob Pen

Oct 18, 2020 19:12

Let us say $\frac{1}{0} = x$ but $1 \ne 0*x$

Bob Pen

Oct 18, 2020 19:10

That is possible, but I don't really know the correct answer to why that is allowed and dividing by zero is not allowed

Bob Pen

Oct 18, 2020 19:08

Because there is nothing to divide on

Bob Pen

Oct 18, 2020 19:08

It just doesn't make sense?

Bob Pen

Oct 18, 2020 19:06

It means that the function is undefined at that point?

Bob Pen

Oct 18, 2020 19:05

That is a good point

Bob Pen

Oct 18, 2020 19:04

If I insert $a$ for $x$ I will get $\frac{1}{0}$ which is not possible ( I think )

Bob Pen

Oct 18, 2020 19:01

That I do

Bob Pen

Oct 18, 2020 18:58

Which means that the limit I have got will give a division by zero, which is not allowed

Bob Pen

Oct 18, 2020 18:57

if x approaches a

Bob Pen

Oct 18, 2020 18:57

Wouldn't it be zero?

Bob Pen

Oct 18, 2020 18:55

Really lost on this one, sorry

Bob Pen

Oct 18, 2020 18:54

Not defined?

Bob Pen

Oct 18, 2020 18:51

I was thinking L'Hopital but that wouldn't get me very far

Bob Pen

Oct 18, 2020 18:49

Found a task in my Professors exercises which I am not sure I understand:
Find $lim_{x \rightarrow a+0} \frac{1}{(x-a)^{2k}} \, \, \, k \in \mathbb{N}$
How would one proceed with such a task?

Sonofgreek

Oct 1, 2020 10:54

math.stackexchange.com/questions/1728376/… Trying to solve exactly the same problem as this one. I defined $\delta = \frac{1}{2}$, which means that $|x| < 1/2$. And $|x^2 - 1| > \frac{3}{4}$. But how does one proceed from this to create the relationship between $\delta$ and $\epsilon$?

Sonofgreek

Sep 29, 2020 21:15

Sorry, if I am asking very obvious questions

Sonofgreek

Sep 29, 2020 21:14

I think I might just be stupid, would this mean that $s$ could have been $\frac{a}{a^2-2b^2}$ and $t$ could have been $\frac{b}{a^2-2b^2}$. And as long as $s,t$ were still rational numbers, they would be in $S$?

Sonofgreek

Sep 29, 2020 21:00

I see! And then because of how S is defined it would $s-t\sqrt(2)$ part will show that it is in S?

Sonofgreek

Sep 29, 2020 20:49

Hello everyone! Suppose I have a set defined by $S= {s + t\sqrt{2} \, \, | s,t \in \mathbb{Q}}$ And $x \neq 0, x \in S$, how can I show that $\frac{1}{x} \in S$ as well? Any guideline or tips would be very much appreciated.

Sonofgreek

May 19, 2020 14:11

Can type it in MathJax; $x- 2y + 3z =1$

Sonofgreek

May 19, 2020 14:08

Hey! Quick question is x-2y+3z=1 a subspace of R^3?

$Mathematics$ Calvin's Chat Room

A place to sometimes discuss mathematics, fluids, mathematical...

Sonofgreek

Oct 1, 2020 13:04

Still stuck, can’t seem to wrap my head around it

Sonofgreek

Oct 1, 2020 11:54

Any further advice? Sorry for the stupidity

Sonofgreek

Oct 1, 2020 11:53

If I I set $4x^2/3 \le \epsilon$ I get $x \le \sqrt{3 \epsilon} / 2$

Sonofgreek

Oct 1, 2020 11:52

Hmm

Sonofgreek

Oct 1, 2020 11:25

Sorry if I am asking very obvious questions, I am having a hard time understanding Andre's suggestion

Sonofgreek

Oct 1, 2020 11:25

I tried making sense of it so, $|x^2-1| > 3/4$, which means that $\frac{x^2}{x^2-1} \lt \frac{4x^2}{3}$ which in turn means that $|f(x) - L| \le \frac{4x^2}{3}$. But how do I go forward from here?

Sonofgreek

Oct 1, 2020 11:20

However, I am struggling to proceed from this. Do you have any advice?

Sonofgreek

Oct 1, 2020 11:19

Ah, that makes sense. To limit $\delta$

Sonofgreek

Oct 1, 2020 11:17

https://math.stackexchange.com/questions/1728376/proof-limit-of-frac1x2-1-1-using-epsilon-delta-as-x-0-t Trying to solve exactly the same problem as this one. I defined $\delta = \frac{1}{2}$, which means that $|x| < 1/2$. And $|x^2 - 1| > \frac{3}{4}$. But how does one proceed from this to create the relationship between $\delta$ and $\epsilon$?