# Symmetric positive definite matrix

rendinat

## 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|}≤$\frac{1}{2}$max{aii}+$\frac{1}{2}$max{ajj}

then I am stuck! :(

## The Attempt at a Solution

Homework Helper
Gold Member
Try applying ##a_{ii} \leq |a_{ii}| \leq \max |a_{ii}|## to the original inequality.

rendinat
But how does that help with the equality I need?

Homework Helper
Dearly Missed

## 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|}≤$\frac{1}{2}$max{aii}+$\frac{1}{2}$max{ajj}

then I am stuck! :(

## 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
$$Q(x) = x^T A x = \sum_i a_{ii} x_i^2 + 2 \sum_{i < j} a_{ij} x_i x_j$$
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
$$Q(\tilde{x}) = a_{11} x_1^2 + 2 a_{12} x_1 x_2 + a_{22}x_2^2.$$ 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##.

rendinat
By proving that it is not greater in that instance still doesn't prove they are equal, does it?

Homework Helper
Dearly Missed
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.

rendinat
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?

Homework Helper
Gold Member
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##?

rendinat
|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|}≤$\frac{1}{2}$max{aii}+$\frac{1}{2}$max{ajj}

Homework Helper
Gold Member
So I shouldn't do the steps I did to get to? max{|aij|}≤$\frac{1}{2}$max{aii}+$\frac{1}{2}$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}|##?

rendinat
Aren't they equal and I get that max{|aij|}≤max{aii}

Homework Helper
Gold Member
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?

rendinat
That has been my question, how can I show/prove that they are equal?

Homework Helper
Gold Member
If you can show the opposite inequality, that will suffice, right? In other words, is it true that
$$\max |a_{ii}| \leq \max |a_{ij}|$$

Homework Helper
Dearly Missed
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##.