Questions regarding minors of a positive definite matrix In my lectures on Matrix computations, there's a section titled Gaussian elimination and Cholesky decomposition. It is a follows:
suppose $A=A^T$ is positive definite, $a_{11>0}$ and $A_{22}$ is positive definite as well, then we have:
$
A=\begin{bmatrix}
a_{11} & A_{21}^T\\
A_{21} & A_{22}\\
\end{bmatrix}
$
$
E_{1}=\begin{bmatrix}
1 & 0\\
-A_{21}.a_{11}^{-1} & I_{n-1}\\
\end{bmatrix}
$
$\implies$
$
E_1A=\begin{bmatrix}
a_{11}&A_{21}^T\\
0&A_{22}-A_{21}.a_{11}^{-1}.A_{21}^T
\end{bmatrix}
=A^{(2)}
$
I don't seem to understand why $A_{22}$ has to be positive definite. And from there, why is $A_{22}-A_{21}.a_{11}^{-1}.A_{21}^T$ positive definite, I don't get it. 
 A: For the other expression see that for $u = (x, Y^T)^T \neq 0$ where $Y\in \mathbb R^{n-1}$ and $x\in\mathbb R$ we have
$$\begin{align*}u^T A u & = (x\quad Y^T) \left(\begin{matrix} xa_{11}+ A_{21}^T Y \\ xA_{21} + A_{22} Y\end{matrix}\right) \\
& = x^2a_{11} + x A_{21}^T Y + Y^T A_{21} x + Y^T A_{22} Y \\
& =x^2a_{11} + 2xA_{21}^T Y + Y^T A_{22} Y > 0 \end{align*}$$
If you chose $x = 0$ and $Y \neq 0$ you have $A_{22} \succ 0$.  

For $x = -A_{21}^TY \cdot a_{11}^{-1}$ and $Y \neq 0$ we get
$$\begin{align*}
\underbrace{Y^T A_{21} a_{11}^{-2} A_{21}^T Y}_{=x^2} \cdot a_{11} \underbrace{-2a_{11}^{-1} (A_{21}^TY)^2}_{=2 x A_{21}^T Y} + Y^T A Y & > 0 \\
-a_{11}^{-1} (A_{21}^TY)^2 + Y^T A_{22} Y & > 0 \\
Y^T (A_{22} - a_{11}^{-1} A_{21} A_{21}^T)Y & > 0\end{align*}$$
And you're done.
A: Let $A$ be a $n\times n$ symmetric matrix. Then $A$ is positive definite if and only if for all non-zero $\mathbf{x}$ in $\mathbb{R}^n$, we have
$$\mathbf{x}^\mathrm{T}A\mathbf{x} > 0$$
If we take the first entry of $\mathbf{x}$ to be zero, then we have through block multiplication
$$0 < \mathbf{x}^\mathrm{T}A\mathbf{x} = \begin{pmatrix}0 & \mathbf{x}_{n-1}^\mathrm{T}\end{pmatrix}\begin{pmatrix}a_{11} & \mathbf{a}_{12}^\mathrm{T} \\ \mathbf{a}_{12} & A_{22}\end{pmatrix}\begin{pmatrix}0 \\ \mathbf{y}_{n-1}\end{pmatrix}=\mathbf{x}_{n-1}^\mathrm{T} A_{22} \mathbf{x}_{n-1}$$
Since $\mathbf{x}$ was arbitrary, it follows that its last $n-1$ components are also arbitrary. So it holds that
$$\mathbf{x}_{n-1}^\mathrm{T}A_{22}\mathbf{x}_{n-1} > 0$$
for all non-zero $\mathbf{x}_{n-1}$ in $\mathbb{R}^{n-1}$. By definition, this means that $A_{22}$ is positive-definite.
To see the second part, it helps to multiply the result by the transpose of $M_1$. Then we end up with
$$M_1AM_1^\mathrm{T} = \begin{pmatrix}a_{11} & \mathbf{0}^\mathrm{T} \\ \mathbf{0} & A_{22}-a_{11}^{-1}A_{12}A_{12}^\mathrm{T}\end{pmatrix}$$
Notice that $M_1AM_1^\mathrm{T}$ is symmetric. Furthermore, it is positive-definite since
$$\mathbf{x}^\mathrm{T}M_1AM_1^\mathrm{T}\mathbf{x} = (M_1^\mathrm{T}\mathbf{x})^\mathrm{T}A(M_1^\mathrm{T}\mathbf{x}) > 0$$
so that positive-definiteness is inherited from $A$. Formally, this corresponds to a change of basis for $A$ which does not change the underlying billinear form (thus remaining positive-definite). Note that we have implicitly used the fact that $M_1$ is invertible so that $M_1^\mathrm{T}\mathbf{x} \neq \mathbf{0}$ for $\mathbf{x} \neq \mathbf{0}$.
Therefore, from the previous result, it follows that the submatrix $A_{22}-a_{11}^{-1}A_{12}A_{12}^\mathrm{T}$ is also positive-definite.
