# How do the values of $a$ affect definiteness of this matrix $\textbf{A}$?

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?

• The determinant should be 2a^2-6 – Alejandro Menaya Apr 24 at 7:21
• I don't know if I should feel released or more worried from the fact I cometed such a silly arithmetic mistake. Thank you. – torito verdejo Apr 24 at 7:30

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