How to find eigenvalues $\lambda>0$ so that matrix A is positive and definite We are given matrix A:
\begin{pmatrix} 
  s    & -1 & -1\\ 
  -1 & s & -1\\
-1&-1&s\\ 
\end{pmatrix}
I need to find for which s do A has all eigenvalue $\lambda>0$(positive definite).
The main problem is that i can try with different values but i will never found all of them.
Is there a trick to do this?
 A: The eigenvalues $\lambda$ of $A$ are given by:
$$\det{\left(A-\lambda I\right)}=0\Longleftrightarrow\begin{vmatrix}s-\lambda&-1&-1\\-1&s-\lambda&-1\\-1&-1&s-\lambda\end{vmatrix}=0$$
Expanding the determinant and simplifying a bit, you get the following characteristic equation:
$$(\lambda-s-1)^2(\lambda-s+2)=0$$
Therefore, the eigenvalues are $\lambda=s+1(\text{twice}),s-2$. Thus, if you want $A$ to be positive definite:
$$\lambda>0\Longrightarrow s-2>0\Longleftrightarrow s>2$$
A: Sylvester's criterion tells you that the conditions are
\begin{align}
&\det\begin{pmatrix} s \end{pmatrix}>0\\[12px]
&\det\begin{pmatrix} s & -1 \\ -1 & s\end{pmatrix}>0\\[12px]
&\det\begin{pmatrix} 
  s    & -1 & -1\\
  -1 & s & -1\\
-1&-1&s\\ 
\end{pmatrix}>0
\end{align}
This means $s>0$, $s^2-1>0$, $s^3-3s-2=0$. The third condition reads $(s+1)^2(s-2)>0$, so the final solution is $s>2$.
A: there is no need to compute the characteristic polynomial.
the reason is that this matrix is a rank one peroration of a scalar matrix $(s+1)I.$  the eigenvalues of $uu^\top$ where $u^\top = (1,1,1)$ are $u^\top u = 3, 0, 0.$ therefore the eigenvalues of $$A = (s+1)I - uu^\top $$ are $s-2, s, s.$ for $A$ to be positive definite, you need $s > 2.$
