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Let $a$ be a real number such that a matrix $\begin{pmatrix} 1 & a & a \\ a & 1 & a \\ a & a & 1 \end{pmatrix}$ has three eigenvalues $\lambda_1 \geq \lambda_2 \geq \lambda_3 > 0$. The interval of $a$ must be $\cdots$

I have applied the definitions of eigenvalues and stuck at its characteristic polynomial form. I have $(1-\lambda)^3 + 2a^3 - 3a^2(1-\lambda) = 0$ Based on the equation, I learn that $a = 0$ make the eigenvalues of the matrix is only one, that is $1$, thus satisfied the conditions. But, what is the interval of $a$? Do you have another idea?

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  • $\begingroup$ Write the equation as a polynomial in $\lambda$. It is a polynomial of degree $3$. You can write its roots as functions of $a$. $\endgroup$ – Levent Mar 11 '18 at 14:25
  • $\begingroup$ The problem statement is awkward. This matrix has at most two distinct eigenvalues. You’re meant to find the values of $a$ for which all the eigenvalues are positive. $\endgroup$ – amd Mar 11 '18 at 15:49
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Replace $a=1-b$.

Then for $(1-\lambda)^3+2(1-b)^3-3(1-b)^2(1-\lambda)=0$ you can notice that $\lambda=b$ is an evident root.

This helps you factorize the degree $3$ characteristic polynomial and find eigenvalues $\{3-2b,b,b\}$.

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  • $\begingroup$ Where did you that set of eigenvalues? $\endgroup$ – Shane Dizzy Sukardy Mar 11 '18 at 15:14
  • $\begingroup$ These are the roots of the polynomial of degree $3$ in $\lambda$ above. Since $b$ is evident root, factorize by $(\lambda-b)$ and so on.Nothing mysterious, just factorize the polynomial to $(\lambda-b)^2(\lambda+2b-3)$ $\endgroup$ – zwim Mar 11 '18 at 16:09

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