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Let $$\textbf{A} = \begin{bmatrix} -2 & 0 & 1 \\ 0 & -2 & a \\ 1 & a & -2\end{bmatrix}$$

Given that one of its eigenvalues is equal to $-2$, how does its definiteness vary with $a$?

From the fact that there is a negative eigenvalue, I know that $\textbf{A}$ cannot be positive (semi-)definite, and since its top left principal minors $$|\textbf{A}_1| = -2<0, |\textbf{A}_2| = 4 >0$$ I expect to find how its definiteness depends on $a$ by the solution to this equation $$|\textbf{A}| =2a^2-4= 0 \iff a = \pm \sqrt{2}$$ which would imply that $\textbf{A}$ is negative definite or negative semi-definite or indefinite whether $a \in ]-\sqrt{2}, \sqrt{2}\ [$ or $a =\pm\sqrt{2}$ or $a \notin [-\sqrt{2}, \sqrt{2}\ ]$ respectively.

Nevertheless, this is not the answer I'm given, which is similar to mine, but with $\pm \sqrt{3}$ instead. Why is my reasoning wrong, and how do we get to the proper answer?

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    $\begingroup$ The determinant should be 2a^2-6 $\endgroup$ Apr 24, 2019 at 7:21
  • $\begingroup$ I don't know if I should feel released or more worried from the fact I cometed such a silly arithmetic mistake. Thank you. $\endgroup$
    – Albert
    Apr 24, 2019 at 7:30

1 Answer 1

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I realized, thanks to the comment of @AlejandroMenaya that my mistake was due to a simple arithmetic error, i.e. $|\textbf{A}| = 2a^2 -6$.

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