For $n \in \mathbb{N}^*$, $A \in \mathcal{S}_n(\mathbb{R})$ and $i \in \llbracket 1; n \rrbracket$, we denote by $A^{(i)}$ the square matrix of order $i$ extracted from $A$, consisting of the first $i$ rows and the first $i$ columns of $A$. For all $n \in \mathbb{N}^*$, we say that a matrix $A$ of $\mathcal{S}_n(\mathbb{R})$ satisfies property $\mathcal{P}_n$ if $\operatorname{det}\left(A^{(i)}\right) > 0$ for all $i \in \llbracket 1; n \rrbracket$. In the special cases $n = 1$ and $n = 2$, show directly that any matrix $A \in \mathcal{S}_n(\mathbb{R})$ satisfying property $\mathcal{P}_n$ is positive definite.
For $n \in \mathbb{N}^*$, $A \in \mathcal{S}_n(\mathbb{R})$ and $i \in \llbracket 1; n \rrbracket$, we denote by $A^{(i)}$ the square matrix of order $i$ extracted from $A$, consisting of the first $i$ rows and the first $i$ columns of $A$. For all $n \in \mathbb{N}^*$, we say that a matrix $A$ of $\mathcal{S}_n(\mathbb{R})$ satisfies property $\mathcal{P}_n$ if $\operatorname{det}\left(A^{(i)}\right) > 0$ for all $i \in \llbracket 1; n \rrbracket$.
In the special cases $n = 1$ and $n = 2$, show directly that any matrix $A \in \mathcal{S}_n(\mathbb{R})$ satisfying property $\mathcal{P}_n$ is positive definite.