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I am trying to understand the a proof of the Sylvesters Criterion. In this proof we basically show that the a Hermitian matrix $A$ with only positive principal minors cannot be other than positive definite.

The proof starts by supposing that $A$ is non positive definite (PD) and then moves on by taking a linear combination of eigenvectors $v_{1,2}$ (for which $\lambda_{1,2}<0$) s.t. the last entry of it is $0$. Then using the definition of definiteness:

$v^TAv = \alpha^2v_1^TAv_1 + \beta^2v_2^TAv_2 = \alpha^2\lambda_1v_1 + \beta^2\lambda_2v_2<0$

After this point, the author concludes that the submatrix of $A [n-1 \times n-1]$ cannot be a positive minor, which contradicts the setup.

Hence the leading (n−1)×(n−1) principal submatrix of A is not positive definitive.

Thanks in advance

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Pablo Jeken Rico
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Say $v$ is the linear combination of $v_1$ and $v_2$ which has a $0$ in the last coordinate (but, crucially, not everywhere). Let:

  • $w$ be the $(n-1)$-dimensional vector obtained by dropping the last coordinate of $v$.
  • $B$ be the $(n-1) \times (n-1)$ principal submatrix of $A$.

Then $v^T A v = w^T B w$: in $v^TAv$, every entry of $A$ that's also in $B$ appears with the same coefficient as in $w^TAw$, and evry entry of $A$ that's not in $B$ appears with the same coefficient of $0$.

We've shown $v^T A v < 0$, and therefore $w^T B w < 0$, which proves that $B$ is not positive definite.

You can also see my write-up of the same proof in these lecture notes, if you're interested.

Misha Lavrov
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