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This is a homework problem: Let $a,b,c$ be positive real numbers such that $b^2+c^2<a<1$. If $A=\begin{pmatrix} 1&b&c\\b&a&0\\c&0&1\end{pmatrix}$, then which of the following is correct?

  1. All the eigenvalues of $A$ are negative real numbers.
  2. All the eigenvalues of $A$ are positive real numbers.
  3. $A$ has a positive and a negative eigenvalue.
  4. $A$ has a non-real complex eigenvalue.

Since the matrix $A$ is real symmetric so, all the eigenvalues must be real. Thus 4 is wrong. Trace of $A=2+a$ and det $A>0$. Thus 1 is wrong. Now all the eigenvalues may be positive or two of them are negative and one is positive. Please give a hint to understand whether option 2 or 3 is correct.

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Hint: by Sylvester's crieterion, it is enough to check the principal minors. Note that $a - b^2 > 0$.

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  • $\begingroup$ Since all the principal minors are positive, so the matrix is positive definite. $\endgroup$ – Anupam Jan 25 '15 at 2:22

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