Symmetric positive definite matrix

[rendinat]

Homework Statement

In a symmetric positive definite matrix, why does max{|a_ij|}=max{|a_ii|}

Homework Equations

|a_ij|≤(a_ii+a_jj)/2

The Attempt at a Solution

max{|a_ij|}≤max{(a_ii+a_jj)/2

max{|a_ij|}≤max{a_ii/2}+max{a_jj/2}

max{|a_ij|}≤\frac{1}{2}max{a_ii}+\frac{1}{2}max{a_jj}

then I am stuck! :(

 

[jbunniii]

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?

 

[Ray Vickson]

Homework Statement

In a symmetric positive definite matrix, why does max{|a_ij|}=max{|a_ii|}

Homework Equations

|a_ij|≤(a_ii+a_jj)/2

The Attempt at a Solution

max{|a_ij|}≤max{(a_ii+a_jj)/2

max{|a_ij|}≤max{a_ii/2}+max{a_jj/2}

max{|a_ij|}≤\frac{1}{2}max{a_ii}+\frac{1}{2}max{a_jj}

then I am stuck! :(

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?

 

[Ray Vickson]

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{|a_ij|} = max{|a_ii|}

Maybe I am going about it incorrectly and that is the problem. Where would you begin to try to prove these are equal?

 

[jbunniii]

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]

|a_ij|≤(a_ii +a_jj)/2≤(max{a_ii} +a_jj)/2 Is this what you mean?

So I shouldn't do the steps I did to get to? max{|a_ij|}≤\frac{1}{2}max{aii}+\frac{1}{2}max{ajj}

 

[jbunniii]

So I shouldn't do the steps I did to get to? max{|a_ij|}≤\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{|a_ij|}≤max{a_ii}

 

[jbunniii]

Aren't they equal and I get that max{|a_ij|}≤max{a_ii}

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?

 

[jbunniii]

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

 

[Ray Vickson]

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 $i$.

 

[jbunniii]

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.