<s>[INST] Prove if $A$ is positive definite, then $A_{11}$,$A_{22}$ are positive definite. Here ${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
  {{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\ 
  {{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}} 
\end{array}} \right)$. 
I can prove $\mathbf{A}_{11}$ is positive definite but need help with $\mathbf{A}_{22}$. By Sylvester’s criterion, $|\mathbf{A}_{11} |>0$ since $|\mathbf{A}_{11} |$ is a principle minor. Thus $\mathbf{A}_{11}$ is invertible and we have
${\mathbf{A}} = \left( {\begin{array}{*{20}{c}}
  {{{\mathbf{A}}_{11}}}&{{{\mathbf{A}}_{12}}} \\ 
  {{{\mathbf{A}}_{21}}}&{{{\mathbf{A}}_{22}}} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\mathbf{I}}&{\mathbf{O}} \\ 
  {{{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}}&{\mathbf{I}} 
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\ 
  {\mathbf{O}}&{\mathbf{S}} 
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {\mathbf{I}}&{{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}} \\ 
  {\mathbf{O}}&{\mathbf{I}} 
\end{array}} \right)$
where ${\mathbf{S}} = {{\mathbf{A}}_{22}} - {{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}$ is the Schur complement. Since
${\left( {\begin{array}{*{20}{c}}
  {\mathbf{I}}&{\mathbf{O}} \\ 
  {{{\mathbf{A}}_{21}}{\mathbf{A}}_{11}^{ - 1}}&{\mathbf{I}} 
\end{array}} \right)^{\text{T}}} = \left( {\begin{array}{*{20}{c}}
  {\mathbf{I}}&{{{\left( {{\mathbf{A}}_{11}^{\text{T}}} \right)}^{ - 1}}{\mathbf{A}}_{21}^{\text{T}}} \\ 
  {\mathbf{O}}&{\mathbf{I}} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\mathbf{I}}&{{\mathbf{A}}_{11}^{ - 1}{{\mathbf{A}}_{12}}} \\ 
  {\mathbf{O}}&{\mathbf{I}} 
\end{array}} \right)$
we have $\mathbf{A}$ is congruent with $\left( {\begin{array}{*{20}{c}}
  {{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\ 
  {\mathbf{O}}&{\mathbf{S}} 
\end{array}} \right)$. If $$ is positive definite, then $\left( {\begin{array}{*{20}{c}}
  {{{\mathbf{A}}_{11}}}&{\mathbf{O}} \\ 
  {\mathbf{O}}&{\mathbf{S}} 
\end{array}} \right)$ is positive definite, and $_{11}$,$\mathbf{S}$ are both positive definite, since a block diagonal matrix is positive definite iff each diagonal block is positive definite.
Anyone can help continue to prove $_{22}$ is also positive definite? Thank you!
 [/INST] To elaborate on Stella's answer, let's say $A_{11}$ is $5\times5$. Then if you take $$x=\begin{bmatrix}a \\ b \\ c \\ d \\e \\0 \\ \cdots \\ 0 \end{bmatrix}$$ then $$x^T A x = \begin{bmatrix}a & b & c & d & e \end{bmatrix}A_{11}\begin{bmatrix}a \\ b \\ c \\ d \\e \end{bmatrix}$$ and so if $x^T A x$ is positive for all nontrivial x then the product with $A_{11}$ must be positive for all nontrivial $a,b,c,d,e$ which shows that $A_{11}$ is positive definite.
</s>