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.