# Proof about a positive definite matrix

• I
• Jamister

#### Jamister

TL;DR Summary
A symmetric real matrix ##A## with positive elements ##a_{i,j}\geq 0## can’t be definite positive matrix
I need to prove the following:

A symmetric real matrix ##A## with positive elements ##a_{i,j}\geq 0## can’t be definite positive matrix (i.e. with only positive eigenvalues) if the following condition holds:
$$\sum_{i=1}^{N-1}a_{i,i+1}>\frac{1}{2}\sum_{i=1}^{N}a_{i,i}=\frac{1}{2}\text{Tr}(A)$$
and in addition the only non zero elements ##a_{i,j}## are those that ## i-1 \leq j \leq i+1##
Does anyone have ideas?

Last edited:

Summary: A symmetric real matrix ##A## with positive elements ##a_{i,j}\geq 0## can’t be definite positive matrix

I need to prove the following:

A symmetric real matrix ##A## with positive elements ##a_{i,j}\geq 0## can’t be definite positive matrix (i.e. with only positive eigenvalues) if the following condition holds:
$$\sum_{i=1}^{N-1}a_{i,i+1}>\frac{1}{2}\sum_{i=1}^{N}a_{i,i}=\frac{1}{2}\text{Tr}(A)$$
Does anyone have ideas?
As per forum rules, you need to show your own efforts. So, what have you tried?

swampwiz and Jamister
I tried many things...
1. using the characteristic polynomial to show that there are negative roots
2. using Sylvester's criterion to show the matrix is not a positive definite
3. using the inequalities ##\left|m_{i j}\right| \leq \sqrt{m_{i i} m_{j j}} \quad \forall i, j## (and it turns out these are not sufficient to show the claim, provided by counterexamples)
4. using Cholesky decomposition.

I added another condition to the question.

Can you think of a vector x such that ##x^T A x < 0##?

Jamister
Can you think of a vector x such that ##x^T A x < 0##?
I don't know...

Try restricting to 2x2 matrices and see if you can find a vector.

yes 2x2 is easy I can find. the eigenvector $$(1,-1)$$

I want to edit the post. why It seems like I can't do it...

IIRC there is a time limit for editing posts unless you have additional privileges. For example, I can edit any of my posts but I am not completely sure which of my shiny badges is responsible

Jamister
I want to say that the matrix is tridiagonal
IIRC there is a time limit for editing posts unless you have additional privileges. For example, I can edit any of my posts but I am not completely sure which of my shiny badges is responsible

Either way, what about a 3x3 matrix? Can you find a vector ##x## such that ##x^T A x## contains the relevant sums?

Jamister
Either way, what about a 3x3 matrix? Can you find a vector ##x## such that ##x^T A x## contains the relevant sums?
great! I find a way and proved it! thank you!

SammyS