• Support PF! Buy your school textbooks, materials and every day products Here!

Symmetric positive definite matrix

  • Thread starter rendinat
  • Start date
  • #1
8
0

Homework Statement



In a symmetric positive definite matrix, why does max{|aij|}=max{|aii|}

Homework Equations



|aij|≤(aii+ajj)/2

The Attempt at a Solution



max{|aij|}≤max{(aii+ajj)/2

max{|aij|}≤max{aii/2}+max{ajj/2}

max{|aij|}≤[itex]\frac{1}{2}[/itex]max{aii}+[itex]\frac{1}{2}[/itex]max{ajj}

then I am stuck! :(

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
Try applying ##a_{ii} \leq |a_{ii}| \leq \max |a_{ii}|## to the original inequality.
 
  • #3
8
0
But how does that help with the equality I need?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement



In a symmetric positive definite matrix, why does max{|aij|}=max{|aii|}

Homework Equations



|aij|≤(aii+ajj)/2

The Attempt at a Solution



max{|aij|}≤max{(aii+ajj)/2

max{|aij|}≤max{aii/2}+max{ajj/2}

max{|aij|}≤[itex]\frac{1}{2}[/itex]max{aii}+[itex]\frac{1}{2}[/itex]max{ajj}

then I am stuck! :(

Homework Statement





Homework Equations





The Attempt at a Solution

Assume that the word "symmetric" is always present in the following. Remember that a positive-definite matrix is, by definition, one that renders the quadratic form
[tex] Q(x) = x^T A x = \sum_i a_{ii} x_i^2 + 2 \sum_{i < j} a_{ij} x_i x_j [/tex]
positive for all vectors ##x## that are not the zero vector. You are being asked to show that for a positive-definite matrix, the largest element occurs on the diagonal (in terms of magnitudes, that is). So, assume the contrary: that the largest element is not on the diagonal. By re-labelling if necessary we can assume that ##|a_{12}|## is the largest element. Now look at vectors of the form ##\tilde{x} = (x_1, x_2, 0, 0, \ldots,0)^T##, which give
[tex] Q(\tilde{x}) = a_{11} x_1^2 + 2 a_{12} x_1 x_2 + a_{22}x_2^2.[/tex] Now try to show that if ##|a_{12}| > a_{11}## and ##|a_{12}| > a_{22}## then we can find ##(x_1,x_2) \neq (0,0)## that makes ##Q(\tilde{x}) < 0##.
 
  • #5
8
0
By proving that it is not greater in that instance still doesn't prove they are equal, does it?
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
By proving that it is not greater in that instance still doesn't prove they are equal, does it?
I have no idea what you are talking about. I am just trying to derive a contradiction by assuming an off-diagonal element is maximal.
 
  • #7
8
0
I am trying to prove that max{|aij|} = max{|aii|}

Maybe I am going about it incorrectly and that is the problem. Where would you begin to try to prove these are equal?
 
  • #8
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
But how does that help with the equality I need?
I can't do the problem for you. What do you get when you plug in the fact that ##a_{ii} \leq \max |a_{ii}|## into ##|a_{ij}| \leq (a_{ii} + a_{jj}) / 2##?
 
  • #9
8
0
|aij|≤(aii +ajj)/2≤(max{aii} +ajj)/2 Is this what you mean?

So I shouldn't do the steps I did to get to? max{|aij|}≤[itex]\frac{1}{2}[/itex]max{aii}+[itex]\frac{1}{2}[/itex]max{ajj}
 
  • #10
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
So I shouldn't do the steps I did to get to? max{|aij|}≤[itex]\frac{1}{2}[/itex]max{aii}+[itex]\frac{1}{2}[/itex]max{ajj}
In fact, it should be ##\max |a_{ij}| \leq (\max |a_{ii}| + \max |a_{jj}|) / 2##.

What can you say about ##\max |a_{ii}|## and ##\max |a_{jj}|##?
 
  • #11
8
0
Aren't they equal and I get that max{|aij|}≤max{aii}
 
  • #12
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
Aren't they equal and I get that max{|aij|}≤max{aii}
You dropped the absolute values on the right hand side. It should be ##\max |a_{ij}| \leq \max |a_{ii}|##. Can you see why this gives you the result you need?
 
  • #13
8
0
That has been my question, how can I show/prove that they are equal?
 
  • #14
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
If you can show the opposite inequality, that will suffice, right? In other words, is it true that
$$\max |a_{ii}| \leq \max |a_{ij}|$$
 
  • #15
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
You dropped the absolute values on the right hand side. It should be ##\max |a_{ij}| \leq \max |a_{ii}|##. Can you see why this gives you the result you need?
No need for absolute values on the right. If A is positive-definite we MUST have ##a_{ii} > 0## for all ##i##. If we weaken the assumption to positive-semidefinite, we still have ##a_{ii} \geq 0## for all ##i##.
 
  • #16
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
No need for absolute values on the right. If A is positive-definite we MUST have ##a_{ii} > 0## for all ##i##. If we weaken the assumption to positive-semidefinite, we still have ##a_{ii} \geq 0## for all ##i##.
True, good point. For this problem, it might be clearer to use the absolute values anyway, even if they are redundant for the diagonal elements.
 

Related Threads for: Symmetric positive definite matrix

  • Last Post
Replies
2
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
708
  • Last Post
Replies
3
Views
5K
Replies
15
Views
5K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
1
Views
1K
Top