Prove that the maximal entries of a positive definite, symmetric, real matrix are on the diagonal

Question

Prove that the maximal entries of a positive definite, symmetric, real matrix are on the diagonal. (Algebra by Artin, Edition 2, chapter 8, problem 2.1)

This is an assignment problem so I don't want a complete solution. I'm not really sure what "maximal" means. Can anyone tell me that?

Answer 1

Since the matrix entries are real, "maximal" here just means largest.

I would interpret the question to mean: given a positive definite real matrix $M$, let $m = \max_{i,j} M_{ij}.$ Prove that whenever $M_{ij} = m$, $i=j$.

Answer 2

Here is a different proof.

We know that the quadratic form associated with such a matrix is

$$Q(X)=X^TMX=\sum m_{ii}x_i^2+2\sum_{i<j} m_{ij}x_ix_j\tag{1}$$

(where $X$ is the column vector $(x_1,x_2,\cdots x_n)^T$).

Let us assume the contrary, i.e., that there is an absolute maximum occurring off diagonal for a certain $m_{ij}$ ($i<j$).

Consider the particular $X=(0...0,1,0,...0,-1,0,...)$ where the 'one' occurs at position $i$ and the 'minus one' occurs at position $j$.

Then expression (1) becomes $m_{ii}+m_{jj}-2m_{ij}<0$.

Contradiction with the definite positiveness.