Definiteness of a symmetric matrix of order $3\times 3$

Question

Let $a,b,c$ be three positive real numbers such that $b^{2}+c^{2}\lt a\lt 1$. Consider $3\times 3$ matrix$$A = \left[ \begin{matrix} 1 & b & c \\ b & a & 0 \\ c & 0 & 1 \end{matrix} \right] $$

Is the matrix positive definite or not?

We know if the quadratic form of $A$ is positive definite then $A$ itself a positive definite. But how we can show that the quadratic form of $A$ ,say, $Q(x)$ is positive definite? Or any other way to show it?

Answer 1

Note that a symmetric matrix is positive definite if and only if the determinants of the upper-left submatrices are all positive. See Sylvester's criterion.

In particular, it is sufficient to check whether $$ 1 > 0\\ a - b^2 > 0\\ (a - b^2) - ac^2 > 0 $$

Answer 2

Hint: $A$ is positive definite if and only if its eigenvalues are (real and) positive. Remember that the eigenvalues are the roots of $\det(A - \lambda I) = 0$.