TITLE: Characterization of positive definite matrix with principal minors
QUESTION [22 upvotes]: A symmetric matrix $A$ is positive definite if $x^TAx>0$ for all $x\not=0$.
However, such matrices can also be characterized by the positivity of the principal minors.
A statement and proof can, for example, be found on wikipedia:
http://en.wikipedia.org/wiki/Sylvester%27s_criterion
However, the proof, as in most books I have seen, is very long and involved. This makes sense in a book where you wanted to prove the other theorems anyway. But there has to be a much better way to prove it.

What is the "proof from the book" that positive definite matrices are characterized by their $n$ positive principal minors?

REPLY [5 votes]: One may also give a simple self-contained proof using Schur complement and matrix congruence. Note that
$$
\pmatrix{P&b\\ b^\ast &c}=\pmatrix{I&0\\ b^\ast P^{-1}&1}\pmatrix{P\\ &c-b^\ast P^{-1}b}\pmatrix{I&P^{-1}b\\ 0&1}\tag{1}
$$
whenever $P$ is invertible. Therefore, if $A$ is any Hermitian matrix such that
$$
\text{all leading principal submatrices of $A$ are invertible,}\tag{2}
$$
then by applying congruence transforms akin to $(1)$ recursively, we get $A=LDL^\ast$ for some real diagonal matrix $D$ and some lower triangular matrix $L$ whose diagonal entries are all equal to $1$. Consequently, if $A_k,D_k$ and $L_k$ denote the leading principal $k\times k$ submatrices of $A,D$ and $L$ respectively, we have $A_k=L_kD_kL_k^\ast$. Hence
$$
\text{$A_k$ is congruent to $D_k$ and $\det(A_k)=\det(D_k)$ for every $k$.}\tag{3}
$$
We are now ready to prove Sylvester's criterion. Suppose $A=A_n>0$. Then $(2)$ is satisfied and $D=D_n>0$ by $(3)$. Hence $\det(A_k)=\det(D_k)>0$ for every $k$.
Conversely, suppose $\det(A_k)>0$ for every $k$. Then $(2)$ is fulfilled and $\det(D_k)=\det(A_k)>0$ by $(3)$ for each $k$, meaning that $D>0$. Hence $A>0$ by $(3)$.