<s>[INST] Positive Definite Matrix Problem Suppose that the Symmetric Matrix
$$B=\left(
\begin{array}{cc}
 \alpha  & a^T \\
 a & A
\end{array}
\right)$$
of order $n+1$ is positive definitive.
(a) Show that the scalar $\alpha$ must be positive and the $n*n$ matrix $A$ must be positive definitive.
(b) What is the Cholesky factorization of $B$ in terms of $\alpha$, $a$, and the Cholesky factorization of $A$?
 [/INST] 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.
</s>