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The sylvester's criterion for positive definite matrices states "A self-adjoint matrix A is positive definite if and only if all the leading principal minors of A have strictly positive determinants".

If we were to focus on the "if" direction (which is more challenging to prove), a possible approach, as suggested in the book "$\textit{Advanced Linear and Matrix Algebra}$" by Nathaniel Johnston, could follow by induction using the fact of such statement being evident on 1x1 matrices (base case) :

For the inductive step, assume that the theorem holds for $(n-1)\times(n-1)$ matrices. To see that it must then hold for $n\times n$ matrices, notice that if $A\in$ $\mathcal{M}_n(\mathbb{F})$ is as in the statement of the theorem and $\det(A_k)>0$ for all $1\leq k\leq n$, then $\det(A)>0$ and (by the inductive hypothesis) $A_{n-1}$ is positive definite. Let $\lambda_i$ and $\lambda_j$ be any two eigenvalues of $A$ with corresponding orthogonal eigenvectors $\mathbf{v}$ and $\mathbf{w}$, respectively. Then define $\mathbf{x}=w_n\mathbf{v}-\nu_n\mathbf{w}$ and notice that $\mathbf{x}\neq\mathbf{0}$ (since $\{\mathbf{v},\mathbf{w}\}$ is linearly independent), but $x_n=w_n\nu_n-\nu_nw_n=0.$ Since $x_n=0$ and $A_{n-1}$ is positive definite, it follows that $$\begin{aligned}\text{0}&<\mathbf{x}^*A\mathbf{x}\\&=(w_n\mathbf{v}-\nu_n\mathbf{w})^*A(w_n\mathbf{v}-\nu_n\mathbf{w})\\&=|w_n|^2\mathbf{v}^*A\mathbf{v}-w_n\overline{\nu_n}\mathbf{w}^*A\mathbf{v}-\nu_n\overline{w_n}\mathbf{v}^*A\mathbf{w}+|\nu_n|\mathbf{w}^*A\mathbf{w}\\&=\lambda_i|w_n|^2\mathbf{v}^*\mathbf{v}-\lambda_iw_n\overline{\nu_n}\mathbf{w}^*\mathbf{v}-\lambda_j\nu_n\overline{w_n}\mathbf{v}^*\mathbf{w}+\lambda_j|\nu_n|^2\mathbf{w}^*\mathbf{w}\\&=\lambda_i|w_n|^2\|\mathbf{v}\|^2-0-0+\lambda_j|\nu_n|^2\|\mathbf{w}\|^2.\end{aligned}$$ Therefore, we can notice that it is not possible that both $\lambda_i\leq0$ and $\lambda_j\leq0.$ Since $\lambda_i$ and $\lambda_j$ were arbitrary eigenvalues of $A$, it follows that $A$ must have at most one non-positive eigenvalue. However, if it had exactly one non-positive eigenvalue then it would be the case that det$(A)=\lambda_1\lambda_2\cdots\lambda_n\leq0$,which we know is not the case. It follows that all of $A’$s eigenvalues are strictly positive, so $A$ is positive definite.

Note: if $\nu_n = w_n =0$ we instead define $\mathbf{x}=\mathbf{v}$ to fix up the proof.

I simply cannot spot where such induction proof would fail if we used it to prove, with small modifications over the inequalities, the statement "If all the leading principal minors of A have non-negative determinants, then A is positive semidefinite". If it indeed is not enough to generalize, then is there a way of "extending" such proof?

excitedGoose
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  • To confirm that there is indeed an issue, the article here provides a 3-by-3 counterexample (Eq. (3)): all the leading principal minors have nonnegative determinants, but the matrix has a negative eigenvalue and thus cannot be PSD. (Alternatively, note that $v^\top Z_1 v=-1$ where $v=(1,0,-1)^\top$.) – Semiclassical Aug 21 '24 at 23:48
  • The author’s proof is not correct. E.g. let ${\mathbf e_1,\mathbf e_2,\mathbf e_3}$ be the standard basis of $\mathbb R^3$ and let $A=I_3$. Take $\mathbf v=\mathbf e_1$ and $\mathbf w=\mathbf e_2$. Then $w_3=v_3=0$. Hence $\mathbf x=w_3\mathbf v-v_3\mathbf w=0$, despite the fact that $\mathbf v$ and $\mathbf w$ are linearly independent. You may see my answer for a correct argument. – user1551 Aug 21 '24 at 23:53
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    @user1551 I added a note that includes how the author handled such edge case. – excitedGoose Aug 22 '24 at 00:05
  • @Semiclassical Indeed, there are practical examples of how the "generalized" statement is false, however the question is where the adapted proof would be failing. – excitedGoose Aug 22 '24 at 00:08
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    Regardless of whether the author’s proof is valid or not, when the leading principal minors are merely nonnegative, it can happen that $v_n=0\ne w_n$ and $\lambda_i=0>\lambda_j$. Thus $\mathbf x^\ast A\mathbf x=0$ although $A$ is not PSD. – user1551 Aug 22 '24 at 00:20
  • As far as extending the criterion to check for positive semi-definiteness goes, you check that all principal minors are $\geq 0$. See https://math.stackexchange.com/questions/4145638/a-is-positive-semidefinite-iff-textdet-b-k-geq-0 – user8675309 Aug 22 '24 at 01:01
  • @user1551 Can't we choose eigenvectors $v,w$ such that $v_nw_n \neq 0$. I guess the argument is: even that's possible, we can have $v_i$ strictly positve while $v_j$ strictly negative such that $\lambda_i|w_n|^2|\mathbf{v}|^2+\lambda_j|\nu_n|^2|\mathbf{w}|^2 = 0$. – GBmath Aug 22 '24 at 05:11
  • @user1551 If instead the statement were to suppose that indeed "All principal minors (not only the leading ones) are non-negative" , does that mean that this proof (with adjusted inequalities) would work once proven that it is not possible to have $v_n=w_n=0$ with $\lambda_i = 0 > \lambda_j$ ? That is, so far, this proof was only good enough to show that there may exist at most one negative eigenvalue along with some extra restrictions ? – excitedGoose Aug 22 '24 at 12:18
  • @excitedGoose How will you leverage the condition "All principal minors are non-negative" in the proof? – GBmath Aug 22 '24 at 16:02
  • The proof would fail in the PSD case. At the end, we would have $\det(A)\ge 0$ and that there would exist at most one strictly negative eigenvalue of $A$. However, $\lambda_1\cdots\lambda_n\le0$ would not contradict $\det(A)\ge 0$. – Paprika7191 Aug 23 '24 at 15:58

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