Why must all the principal sub-matrices' determinants be positive for the matrix to be positive definite? I agree that the eigenvalues has to be positive, but I can't find a connection to the determinants of the principal sub-matrices that I can follow$\dots$
In other words, if $A$ is a symmetrical $n\times n$-matrix that's positive definite, with elements $a_{ii}$. Why must all the principal sub-matrices' determinants be positive (among others $a_{11}*a_{22}-a_{12}*a_{21}>0$)?
 A: Suppose that a symmetric matrix $A=\left(a_{ij}\right)\in {\sf M}_{n\times n}(\mathbb{R})$ is 
positive definite, then given $1\le k\le n$, we consider its principal $k\times k$ submatrix $A_k$. Since for any non-zero vector ${\bf x}\in\mathbb{R}^n$, we have
$${\bf x}^\top A{\bf x}>0.$$
So we let ${\bf x}=(x_1,x_2,\ldots,x_k,0,0\cdots,0)^\top\in\mathbb{R}^n$ be such that the last $n-k$ entries are zero, then we see that
\begin{align}
{\bf x}^\top{A}{\bf x}
&=\begin{pmatrix}
x_1&\cdots&x_k&0&\cdots&0
\end{pmatrix}A
\begin{pmatrix}
x_1\\\vdots\\x_k\\0\\\vdots\\0
\end{pmatrix}\\
&=\begin{pmatrix}
\displaystyle\sum_{i=1}^ka_{i1}x_i
&\cdots&
\displaystyle\sum_{i=1}^ka_{ik}x_i
&0&\cdots&0
\end{pmatrix}
\begin{pmatrix}
x_1\\\vdots\\x_k\\0\\\vdots\\0
\end{pmatrix}\\
&=\sum_{j=1}^k\sum_{i=1}^ka_{ij}x_ix_j,
\end{align}
which is equal to 
\begin{align}
({\bf x}^{(k)})^\top{A_k}{\bf x}^{(k)},
\end{align}
where ${\bf x}^{(k)}\in\mathbb{R}^k$ such that the entries of ${\bf x}^{(k)}$
are the first $k$ entries of ${\bf x}$. Also, it is clear that $A_k$ is symmetric. So we conclude that $A_k$ is also positive definite, and hence $\det(A_k)>0$.
