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?
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 $\{3-2b,b,b\}$.