Find Upper Bound for \| A \| - Homework Statement

  • Thread starter Thread starter tom08
  • Start date Start date
  • Tags Tags
    Bound Upper bound
Click For Summary

Homework Help Overview

The discussion revolves around finding an upper bound \( C \) related to the norm of a matrix \( A \) in the context of vectors \( x \) and \( y \). The original poster seeks to establish a relationship such that \( \| x - Ay \| \leq C \cdot \| x - y \| \).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different expressions for \( C \), including a supremum involving the matrix \( A \) and the identity matrix. There is discussion about the implications of the entries of \( A \) being bounded between 0 and 1, and how this might lead to a more refined estimate for \( C \). Some participants express uncertainty about how to proceed with their reasoning and seek hints or further guidance.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning the assumptions underlying their approaches. Some have provided partial insights into estimating \( C \), while others are looking for clarification or additional hints to advance their understanding.

Contextual Notes

There is an emphasis on the lack of specific information about the matrix \( A \), which affects the ability to derive a more precise upper bound. Participants are also considering the implications of the entries of \( A \) being constrained to a certain range.

tom08
Messages
19
Reaction score
0

Homework Statement



assume that x and y are vectors, and A is a matrix.

can anyone kindly help me to find an upper bound C w.r.t \| A \| s.t.

\| x-Ay \| \leq C \cdot \| x-y\|
 
Physics news on Phys.org
Some quick-and-dirty trial gives me
C = sup( ||(A - I) v|| )
where I is the identity matrix and the supremum is taken over all vectors v.

I wonder if you can do any better, without more information on A.
 
Thank you for ur kind help. if all the entries of A is between 0 and 1, can we get a nicer upper bound ?
 
CompuChip said:
Some quick-and-dirty trial gives me
C = sup( ||(A - I) v|| )
where I is the identity matrix and the supremum is taken over all vectors v.

I wonder if you can do any better, without more information on A.

Could u show me any hints about ur estimate for C.

I only figure out that \|x-Ay\|=\|(x-Ax)+(Ax-Ay)\|\leq\|I-A\| \|x\|+\|A\| \|x-y\|

I don't know how to continue... could anyone kindly give me more hints ?
 
Last edited:
I did
|| x - A y|| = || (x - y) + (y - Ay) ||

But when all entries of A are between 0 and 1, then you can define ||A|| by
|| A || = max(i, j)( |Aij| )
and use that to get a better estimate.
 
CompuChip said:
I did


But when all entries of A are between 0 and 1, then you can define ||A|| by
|| A || = max(i, j)( |Aij| )
and use that to get a better estimate.

if || x - A y|| = || (x - y) + (y - Ay) ||,
then || x - A y|| <= || (x - y) || + || (y - Ay) ||,

but how could u find that C = sup |x-Ax| for all x ?

notice that my esitmate is C*||x-y||
 
can someone give me a hand?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 43 ·
2
Replies
43
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
3K
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K