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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?

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  • $\begingroup$ Have you tried finding the eigenvalues in terms of $s$? (it looks like an ugly cubic, but...) $\endgroup$
    – theage
    May 17 '15 at 13:49
  • $\begingroup$ you have to sove $det(A-\lambda I)=0$, this is a polynomial and you will get something like $(\lambda-\alpha_1)(a\lambda^2+b\lambda+c)$, then you have to check the determinant and play with $s$ to get all $\lambda>0$. $\endgroup$ May 17 '15 at 13:50
  • $\begingroup$ @theage, no it doesn't so ugly, it results $(s-\lambda)^3-2(s-\lambda)-1$ $\endgroup$ May 17 '15 at 13:52
  • $\begingroup$ ...glad to hear it, I guess. $\endgroup$
    – theage
    May 17 '15 at 13:55
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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: $$\lambda>0\Longrightarrow s-2>0\Longleftrightarrow s>2$$

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  • $\begingroup$ @Xenomorph $\ldots=s^3-3s^2\lambda+3s\lambda^2-\lambda^3-3s+3\lambda-2=(-\lambda +s-2)(\lambda^2-2s\lambda-2\lambda+s^2+2s+1)=(-\lambda +s-2)(\lambda^2-2\lambda(s+1)+(s+1)^2)=(-\lambda +s-2)(\lambda -s-1)^2$ $\endgroup$
    – Demosthene
    May 18 '15 at 21:46
  • $\begingroup$ @Xenomorph It's easier if you can guess a root first. $\endgroup$
    – Demosthene
    May 18 '15 at 22:06
  • $\begingroup$ @Xenomorph Yes... so what? $\endgroup$
    – Demosthene
    May 19 '15 at 7:28
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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 $s>2$.

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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 $s > 2.$

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