# Symmetric positive definite matrix

1. Jun 30, 2013

### rendinat

1. The problem statement, all variables and given/known data

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

2. Relevant equations

|aij|≤(aii+ajj)/2

3. 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! :(
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 30, 2013

### jbunniii

Try applying $a_{ii} \leq |a_{ii}| \leq \max |a_{ii}|$ to the original inequality.

3. Jun 30, 2013

### rendinat

But how does that help with the equality I need?

4. Jun 30, 2013

### Ray Vickson

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 $Q(\tilde{x}) < 0$.

5. Jun 30, 2013

### rendinat

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

6. Jun 30, 2013

### Ray Vickson

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. Jun 30, 2013

### 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?

8. Jun 30, 2013

### jbunniii

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. Jun 30, 2013

### 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}

10. Jul 1, 2013

### jbunniii

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. Jul 1, 2013

### rendinat

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

12. Jul 1, 2013

### jbunniii

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. Jul 1, 2013

### rendinat

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

14. Jul 1, 2013

### jbunniii

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

15. Jul 1, 2013

### Ray Vickson

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. Jul 1, 2013

### jbunniii

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.