Prove: symmetric positive definite matrix

Solution 1:

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 $x \in \Bbb R^n$. 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.

Thus, what would happen if $a_{ii}$ was positive?