Characterization of positive definite matrix with principal minors

Question

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?

Answer 1

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.

The forward implication is obvious. If $A$ is positive definite, so are all its leading principal submatrices. Their spectra are hence positive. Thus the leading principal minors are positive, because each of them is a product of the eigenvalues of the submatrix.

To prove the backward implication, we use mathematical induction on $n$. The base case $n=1$ is trivial. Suppose $n\ge2$ and all leading principal minors of $A$ are positive. In particular, $\det(A)>0$. If $A$ is not positive definite, it must possess at least two negative eigenvalues. As $A$ is Hermitian, there exist two mutually 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 definite. By induction assumption, this is impossible. Hence $A$ must be positive definite.

Answer 2

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 $(3)$.

Answer 3

The goal is to study positive definite matrices.

Let ${ A \in \mathbb{C} ^{n \times n} }$ be a Hermitian matrix.

The goal is to express the positive definiteness criteria for ${ A }$ in terms of the entries of ${ A . }$

Say ${ A }$ is positive definite. Setting

$${ x = \begin{pmatrix} x _1 &\cdots &x _k &0 &\cdots &0 \end{pmatrix} ^T \neq 0 , }$$

we have

$${ x ^{\ast} A x = \sum _{i, j = 1} ^n A _{i, j} \overline{x _i} x _j = \sum _{i, j = 1} ^{k} A _{i, j} \overline{x _i} x _j > 0 . }$$

Hence the submatrix

$${ A _k := (A _{i, j}) _{i, j \in [k]} \, \text{ is positive definite}. }$$

Especially all the determinants ${ \det(A _k) > 0 . }$

Now say all the determinants ${ \det(A _k) > 0 , }$ where ${ A _k := (A _{i, j}) _{i, j \in [k]} . }$ Is ${ A }$ positive definite? It turns out yes.

We will proceed by induction on ${ n . }$ The base case ${ n = 1 }$ is clear. Let ${ n \geq 2 . }$ Let

$${ A = \begin{pmatrix} A _{n - 1} &v \\ v ^{\ast} &d \end{pmatrix} }$$

be such that all the determinants ${ \det(A _k) > 0 . }$ By induction hypothesis, ${ A _{n - 1} }$ is positive definite. We are to show

$${ \text{To show: } \quad A = \begin{pmatrix} A _{n - 1} &v \\ v ^{\ast} &d \end{pmatrix} \, \, \, \text{ is positive definite} . }$$

Consider the Schur complement factorisation of ${ A }$ wrt ${ A _{n - 1} , }$ namely

$${ \begin{pmatrix} A _{n - 1} &v \\ v ^{\ast} &d \end{pmatrix} = \begin{pmatrix} I &O \\ v ^{\ast} A _{n -1} ^{-1} &I \end{pmatrix} \begin{pmatrix} A _{n - 1} &O \\ O &d - v ^{\ast} A _{n - 1} ^{-1} v \end{pmatrix} \begin{pmatrix} I &A _{n - 1} ^{-1} v \\ O &I \end{pmatrix} . }$$

It suffices to show

$${ \text{To show: } \quad \begin{pmatrix} A _{n - 1} &O \\ O &d - v ^{\ast} A _{n - 1} ^{-1} v \end{pmatrix} \, \, \, \text{ is positive definite.} }$$

Taking determinants, we have

$${ d - v ^{\ast} A _{n - 1} ^{-1} v > 0 . }$$

Now for ${ x \neq 0 ,}$

$${ {\begin{aligned} &\, x ^{\ast} \begin{pmatrix} A _{n - 1} &O \\ O &d - v ^{\ast} A _{n - 1} ^{-1} v \end{pmatrix} x \\ = &\, \sum _{i, j = 1} ^{n} M _{i, j} \overline{x _i} x _j \\ = &\, \sum _{i, j = 1} ^{n - 1} (A _{n - 1}) _{i, j} \overline{x _i} x _j + \vert x _n \vert ^2 (d - v ^{\ast} A _{n - 1} ^{-1} v ) . \end{aligned}} }$$

If ${ x _n = 0 ,}$ the first term is ${ > 0 }$ and the second term is ${ 0 . }$ If ${ x _n \neq 0 , }$ the first term is ${ \geq 0 }$ and the second term is ${ > 0 . }$

Hence the Schur complement

$${ \begin{pmatrix} A _{n - 1} &O \\ O &d - v ^{\ast} A _{n - 1} ^{-1} v \end{pmatrix} \, \, \, \text{ is positive definite} }$$

as needed. ${ \blacksquare }$