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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

- Proof that $e^{-x} \ge 1-x$
- $p,q,r$ primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.
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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?

- Determinant of a special $n\times n$ matrix
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Sylvester’s criterion says that an $n\times n$ Hermitian matrix $A$ is positive definite if and only if all its leading principal minors of are positive. If one knows that fact that every Hermitian matrix has an orthogonal eigenbasis, one can prove Sylvester’s criterion easily by mathematical induction.

The base case $n=1$ is trivial. The forward implication is also obvious. So, we only need to consider the backward implication.

Suppose all leading principal minors of $A$ are positive. In particular, $\det(A)>0$. It follows that if $A$ is not positive definite, it must possess at least two negative eigenvalues. As $A$ is Hermitian, there exist two *orthogonal* eigenvectors $x$ and $y$ corresponding to two of these negative eigenvalues. Let $u=\alpha x+\beta y\ne0$ be a linear combination of $x$ and $y$ such that the last entry of $u$ is zero. Then $u^\ast Au=|\alpha|^2x^\ast Ax+|\beta|^2y^\ast Ay<0$. Hence the leading $(n-1)\times(n-1)$ principal submatrix of $A$ is not positive definitive. By induction assumption, this is impossible. Hence $A$ must be positive definite.

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