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 $A$?
 A: (a) Since $B$ is positive definite, for every column vector $v$ of order $n+1$, $0<v^TBv$. In particular for $v:=[1, 0, \dots, 0]^T$, we get $0<v^TBv=a$.
Now to show that $A$ is positive definite we let $w$ be any column vector of order $n$ and show that $0<w^TAw$. Set $v:=[0, v_1, v_2, \dots, v_n]^T$ (so $v$ is a column vector of order $n+1$). Since $B$ is positive definite, $0<v^TBv$, but $v^TBv=w^TAw$.
(b) I'm afraid i'm not conversant with the Cholesky factorization, so you're on your own...
A: 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.
A: Given the characterization of a positive definite matrix that says that Its leading principal minors are all positive, if we indicate with $M_n$ the leading principal minor of order $n$ of a generic matrix $M$, we have
\begin{align}
&B_1=\alpha>0\\
&B_k=A_{k-1}>0\qquad\forall k=2,\ldots,n+1
\end{align}
so that also all principal minors of $A$ are positive, and this means that $A$ is a positive definite matrix.
A: $$
\begin{pmatrix}
\sqrt\alpha & 0 \\
\frac{1}{\sqrt\alpha}a & L
\end{pmatrix}
\begin{pmatrix}
\sqrt\alpha & \frac{1}{\sqrt\alpha}a^T \\
0 & L^T
\end{pmatrix} =
\begin{pmatrix}
\alpha & a^T \\
a & LL^T + \frac 1{\alpha}aa^T
\end{pmatrix}
$$
$L$ is obtained from rank-one downdate from the cholesky factor of $A$.
