# For which $a$ is the matrix negative definite?

Find the number of integer numbers of $$a$$ such that the matrix $$\begin{pmatrix}-1& 0 & 5 \\ 0 & a-50 & 0 \\ 5 & 0 & 6-a\end{pmatrix}$$ is negative definite.

For that, the eigenvalues must be negative, right?

I calculated the eigenvalues using Wolfram and we get these.

So all these have to be negative and we get a system of inequalities :

$$a-50<0 \\ \frac{1}{2}\left (-\sqrt{a^2-14a+149}-a+5\right ) <0 \\ \frac{1}{2}\left (\sqrt{a^2-14a+149}-a+5 \right )<0$$ So we get from the first $$a<50$$, from the second $$a\in \mathbb{R}$$ and from the last one $$a>31$$.

So the integer values for $$a$$ are from $$32$$ to $$49$$, so there are $$49-32+1=18$$ values, right?

• I think it is right. But I also think that Sylvester Criterion would be easier here. Commented Feb 5, 2021 at 9:05

$$(\lambda-a+50)(\lambda^2+(a-5)\lambda+a-31)$$
The eigenvalues must all be negative, so, for the first factor, that means $$a<50$$. For the second factor, instead of computing the roots, note that the sum of the roots must be negative, while the product of the roots must be positive. With Vieta's formulas, that means $$a>31$$ and $$a>5$$. So the condition is indeed $$31.