Suppose ${\displaystyle M_{n}}$ is ${\displaystyle n\times n}$  Hermitian matrix ${\displaystyle M_{n}^{\dagger }=M_{n}}$ . Let ${\displaystyle M_{k},k=1,\ldots n}$  be the principal minor matrices, i.e. the ${\displaystyle k\times k}$  upper left corner matrices. It will be shown that if ${\displaystyle M_{n}}$  is positive definite, then the principal minors are positive; that is, ${\displaystyle \det M_{k}>0}$  for all ${\displaystyle k}$ .

${\displaystyle M_{k}}$  is positive definite. Indeed, choosing

${\displaystyle x=\left({\begin{array}{c}x_{1}\\\vdots \\x_{k}\\0\\\vdots \\0\end{array}}\right)=\left({\begin{array}{c}{\vec {x}}\\0\\\vdots \\0\end{array}}\right)}$ 

we can notice that ${\displaystyle 0<x^{\dagger }M_{n}x={\vec {x}}^{\dagger }M_{k}{\vec {x}}.}$  Equivalently, the eigenvalues of ${\displaystyle M_{k}}$  are positive, and this implies that ${\displaystyle \det M_{k}>0}$  since the determinant is the product of the eigenvalues.

To prove the reverse implication, we use induction. The general form of an ${\displaystyle (n+1)\times (n+1)}$  Hermitian matrix is

${\displaystyle M_{n+1}=\left({\begin{array}{cc}M_{n}&{\vec {v}}\\{\vec {v}}^{\dagger }&d\end{array}}\right)\qquad (*)}$ ,

where ${\displaystyle M_{n}}$  is an ${\displaystyle n\times n}$  Hermitian matrix, ${\displaystyle {\vec {v}}}$  is a vector and ${\displaystyle d}$  is a real constant.

Suppose the criterion holds for ${\displaystyle M_{n}}$ . Assuming that all the principal minors of ${\displaystyle M_{n+1}}$  are positive implies that ${\displaystyle \det M_{n+1}>0}$ , ${\displaystyle \det M_{n}>0}$ , and that ${\displaystyle M_{n}}$  is positive definite by the inductive hypothesis. Denote

${\displaystyle x=\left({\begin{array}{c}{\vec {x}}\\x_{n+1}\end{array}}\right)}$ 

then

${\displaystyle x^{\dagger }M_{n+1}x={\vec {x}}^{\dagger }M_{n}{\vec {x}}+x_{n+1}{\vec {x}}^{\dagger }{\vec {v}}+{\bar {x}}_{n+1}{\vec {v}}^{\dagger }{\vec {x}}+d|x_{n+1}|^{2}}$ 

By completing the squares, this last expression is equal to

${\displaystyle ({\vec {x}}^{\dagger }+{\vec {v}}^{\dagger }M_{n}^{-1}{\bar {x}}_{n+1})M_{n}({\vec {x}}+x_{n+1}M_{n}^{-1}{\vec {v}})-|x_{n+1}|^{2}{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}}+d|x_{n+1}|^{2}}$ 

${\displaystyle =({\vec {x}}+{\vec {c}})^{\dagger }M_{n}({\vec {x}}+{\vec {c}})+|x_{n+1}|^{2}(d-{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}})}$ 

where ${\displaystyle {\vec {c}}=x_{n+1}M_{n}^{-1}{\vec {v}}}$  (note that ${\displaystyle M_{n}^{-1}}$  exists because the eigenvalues of ${\displaystyle M_{n}}$  are all positive.) The first term is positive by the inductive hypothesis. We now examine the sign of the second term. By using the block matrix determinant formula

on ${\displaystyle (*)}$  we obtain

${\displaystyle \det M_{n+1}=\det M_{n}(d-{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}})>0}$ , which implies ${\displaystyle d-{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}}>0}$ .

Consequently, ${\displaystyle x^{\dagger }M_{n+1}x>0.}$ 

Let ${\displaystyle M_{n}}$  be an n x n Hermitian matrix. Suppose ${\displaystyle M_{n}}$  is semidefinite. Essentially the same proof as for the case that ${\displaystyle M_{n}}$  is strictly positive definite shows that all principal minors (not necessarily the leading principal minors) are non-negative.

For the reverse implication, it suffices to show that if ${\displaystyle M_{n}}$  has all non-negative principal minors, then for all t>0, all leading principal minors of the Hermitian matrix ${\displaystyle M_{n}+tI_{n}}$  are strictly positive, where ${\displaystyle I_{n}}$  is the nxn identity matrix. Indeed, from the positive definite case, we would know that the matrices ${\displaystyle M_{n}+tI_{n}}$  are strictly positive definite. Since the limit of positive definite matrices is always positive semidefinite, we can take ${\displaystyle t\to 0}$  to conclude.

To show this, let ${\displaystyle M_{k}}$  be the kth leading principal submatrix of ${\displaystyle M_{n}.}$  We know that ${\displaystyle q_{k}(t)=\det(M_{k}+tI_{k})}$  is a polynomial in t, related to the characteristic polynomial ${\displaystyle p_{M_{k}}}$  via

From Minor_(linear_algebra)#Multilinear_algebra_approach, we know that the entries in the matrix expansion of ${\displaystyle \bigwedge ^{j}M_{k}}$  (for j > 0) are just the minors of ${\displaystyle M_{k}.}$  In particular, the diagonal entries are the principal minors of ${\displaystyle M_{k}}$ , which of course are also principal minors of ${\displaystyle M_{n}}$ , and are thus non-negative. Since the trace of a matrix is the sum of the diagonal entries, it follows that

• Gilbert, George T. (1991), "Positive definite matrices and Sylvester's criterion", The American Mathematical Monthly, 98 (1), Mathematical Association of America: 44–46, doi:10.2307/2324036, ISSN 0002-9890, JSTOR 2324036.
• Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN 978-0-521-38632-6. Theorem 7.2.5.
• Carl D. Meyer (June 2000), Matrix Analysis and Applied Linear Algebra, SIAM, ISBN 0-89871-454-0.