I also find a solution for my problem:
$B=\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)$
is a symmetric matrix of order $n+1$ positive definite.
We will show that $\alpha$ is positive and $A$ is a $n\times n$ positive definite matrix.
$B$ is positive definite, so
$\exists\;v=\left( \begin{array}{c} v_1 \\ \tilde{0} \end{array} \right)$
a non-zero $(n+1)\times1$ vector such that $v^TBv$ is positive ($\tilde{0}$ denote $n$ zero elements of $v$).
Consequently, we have
$\left( \begin{array}{cc} v_1 & \tilde{0} \end{array} \right)\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)\left( \begin{array}{c} v_1 \\ \tilde{0} \end{array} \right)=\left( \begin{array}{cc} v_1 \alpha & v_1 a^T \end{array} \right)\left( \begin{array}{c} v_1 \\ \tilde{0} \end{array} \right)=v_1 \alpha v_1=v_1^2\alpha >0$
Because $B$ is positive definite the $v^T B v$ should be positive which in this case forces $\alpha$ to be positive.
Again, considering that
$\exists\;v=\left( \begin{array}{c} 0 \\ \tilde{v} \end{array} \right)$
a non-zero $(n+1)\times1$ vector such that $v^T B v$ is positive ($\tilde{v}$ denote a non-zero $n\times1$ vector).
We can write our formula again:
$\left( \begin{array}{cc} 0 & \tilde{v}^T \end{array} \right)\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)\left( \begin{array}{c} 0 \\ \tilde{v} \end{array} \right)=\left( \begin{array}{cc} \tilde{v}^Ta & \tilde{v}^TA \end{array} \right)\left( \begin{array}{c} 0 \\ \tilde{v} \end{array} \right)=\tilde{v}^TA \tilde{v}>0$
again, because $B$ is positive definite, $v^TB v$ should be positive and it will force $A$ to be positive definite.