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Let $a,b,c$ be a positive real number such that $b^2+c^2<a<1$. Let $A=\begin{bmatrix} 1&b&c\\ b&a & 0\\ c & 0 & 1\end{bmatrix}$. Then

(1) all eigen values of $A$ are positive

(2) all eigenvalues of $A$ are negative

(3) all eigenvalues of $A$ are either positive or negative

(4) all eigenvalues of $A$ are nonreal complex number

Since $A$ is symmetric, all eigen values are real. Hence option (4) is not true.

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    $\begingroup$ Shouldn't (3) be positive or negative? Positive and negative makes it trivially false. $\endgroup$
    – taninamdar
    Sep 25, 2014 at 11:55
  • $\begingroup$ Sorry, i edited $\endgroup$ Sep 25, 2014 at 11:57
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    $\begingroup$ Sylvester's criterion???? $\endgroup$
    – user87543
    Sep 25, 2014 at 12:07
  • $\begingroup$ Just a hint to decide between the first three options, consider the trace. $\endgroup$
    – Macavity
    Sep 25, 2014 at 12:08

1 Answer 1

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Using submatrices: $$\Delta_1=\det\{[1]\}=1>0$$ $$\Delta_2=\det\left\{\left[\begin{array}{cc}1&b\\b&a\end{array}\right]\right\}=a-b^2>a-(b^2+c^2)>0$$ $$\begin{array}{rcl} \Delta_3&=&\det\left\{\left[\begin{array}{ccc}1&b&c\\b&a&0\\c&0&1\end{array}\right]\right\}\\ &=&(a+0+0)-(ac^2+b^2+0)\\ &=&a-ac^2-b^2>a-c^2-b^2>0 \end{array}$$

Then, all subdeterminants are positive, so $A$ is a positive definite matrix and all its eigenvalues are positive.

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