Some Basic Linear Algebra

[Tsakanikas Nickos]

Let \( \displaystyle A \in \mathbb{M}_{n}(\mathbb{R}) \) a symmetric and positive definite matrix. Show that :

1. the diagonal elements of \( \displaystyle A \) are positive.
2. the eigenvalues of \( \displaystyle A \) are positive.
3. the determinant of \( \displaystyle A \) is positive.
4. the absolutely maximum element of \( \displaystyle A \), that is \( \displaystyle \max_{ 1 \leq i,j \leq n } |a_{ij}| \), is on the diagonal, that is \( \displaystyle \max_{ 1 \leq i,j \leq n } |a_{ij}| = a_{kk} , \) for some \( \displaystyle k \in \{ 1, \dots, n\} \).

[Tolaso J Kos]

Here is "half" an answer, since I have my doubts for the first question and I have no solution for the last.

a)
Let \( e_1 = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}\), and so on, where \( e_i \) is a vector of all zeros, except for a \(1 \) in the \( i^{\mathrm{th}} \) place. Since \( A \) is positive definite, then \( x^T A x > 0 \) for any non-zero vector \( x \in \mathbb{ R}^n \). Then, \( e_1^T A e_1 > 0 \), and likewise for \( e_2, \; e_3 \) and so on.

If the \(i^{\mathrm{th}} \) diagonal entry of \(A \) was not positive, \(a_{ii} < 0\), then
$$ e_i^T A e_i = 0\cdot a_{11}\cdot 0 + 1\cdot a_{12}\cdot 0 + \cdots + 1\cdot a_{ii}\cdot 1 + \cdots + 0\cdot a_{nn} \cdot 0$$
since \( e_i \) has zeros everywhere but in the \( i^{\rm th} \) spot.

b) That is basic theory. Since the matrix is positive definite then all eigenvalues are positive.
c) The determinant of the matrix, is actually the product of the eigenvalues that:

$$\det A = \prod_{i} \lambda_i >0$$
where \( \lambda \) is an eigenvalue of \( A \).

I don't have a solution for the last question. May I have a hint?