linear algebra - proofs of property of postive definite and symmetric real matrix Struggled with the following proof regarding properties of real symmetric, positive definite matrix. Tried usage of Sylvester's criterion, but it didn't help. Think that this must be standard result, therefore if anyone would provide link to the proof below, I would be very thankful. Otherwise, I woudl be thankful for advices how to derive proofs for the problem stated below:
If: $A \in  \mathbb{M}_{n\times n}( \mathbb{R})$ is symmetric and positive definite, then:


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*a)$\forall i,j\in $ {$1 ,...,n$}$: i \neq j: a_{i,j}^2 < a_{i,i}a_{j,j}$

*b)$max${$|a_{i,j}|: i,j \in $ {1,...,n} } = max {$ a_{1,1},...a_{n,n}$}
 A: I had some pretty bad typos in my comment. Here are some corrections.

Any submatrix $A_I$ obtained by taking the rows and columns of $A$ indexed by some subset $I\subset \{1,\ldots,n\}$ is positive definite.

For example, $I=\{i,j\}$ shows that $A_I=\begin{bmatrix}a_{ii} & a_{ij}\\a_{ji} & a_{jj}\end{bmatrix}$ is positive definite. Similarly taking $I=\{i\}$ shows that $a_{ii} > 0$.
To see why $A_I$ is positive definite, consider $v^\top A v$ where the only nonzero entries of $v$ are in the components $I$.

Consequently, for any such $A_I$, we have $\det A_I > 0$.

The determinant is the product of the eigenvalues of $A_I$ which are all positive.
Thus a) follows since for $I=\{i,j\}$, $0<\det A_I = a_{ii} a_{jj} - a_{ij}^2$, where the last equality is due to $a_{ii}>0$ as we noted above. This proves b).
This implies $|a_{ij}| < \max(|a_{ii}|,|a_{jj}|)$ for all $i\ne j$, so $\max_{i,j} |a_{ij}| = \max_i |a_{ii}| = \max_i a_{ii}$ (note that diagonal entries are positive, as we showed above), which proves b).
