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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

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?

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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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    $\begingroup$ For posterity's sake, why is the forward implication obvious? $\endgroup$ – Chappers Jun 3 '15 at 16:54
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    $\begingroup$ @Chappers When $A$ is positive definite, every leading principal submatrix of $A$ is positive definite too. $\endgroup$ – user1551 Jun 3 '15 at 17:40
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    $\begingroup$ I understand that that is the implication in question, but why is that obvious? $\endgroup$ – Chappers Jun 3 '15 at 17:49
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    $\begingroup$ @Chappers In the first paragraph, we have taken for granted the fact that every Hermitian matrix is unitarily diagonalisable. So, the determinant of a positive definite matrix, i.e. the product of the eigenvalues, is obviously positive. $\endgroup$ – user1551 Jun 3 '15 at 18:33
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    $\begingroup$ Ah, good, okay. With you now. Thanks. $\endgroup$ – Chappers Jun 3 '15 at 22:05
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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 $(3)$.

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