General structure of a 3 by 3 persymmetric matrix with zero eigenvalues

Here is an interesting problem that might be extended to higher dimensions, I am looking for different simple ways of describing a 3 by 3 matrix with one or two zero eigenvalues. This persymmetric matrix does contain 5 variables that may or may not be equal, that is: $$A=\begin{bmatrix} a_1&a_2&a_3\\ a_4& a_1&a_2\\ a_5&a_4&a_1 \end{bmatrix}.$$ whrere $(a_1,a_2,a_3,a_4,a_5)\in \mathbb{R^5}$. How many independent variables it contains if $A$ has one (or two) zero eigenvalue (by two zero eigenvalues, I mean a zero eigenvalue with the algebraic multiplicity of two).

• (1) By "zero eigenvalues" do you mean the eigenvalues in question are to be actually $0$ (counted with multiplicity)? (2) Are the $a_k$ to be considered as real numbers or as arbitrary complex numbers? – coffeemath May 30 '16 at 23:22
• Both comments are now addressed in the revised question. – Sara Winslet May 30 '16 at 23:34

This Toeplitz matrix has characteristic polynomial $${\lambda}^{3}-3\,a_{{1}}{\lambda}^{2}- \left( -3\,{a_{{1}}}^{2}+2\,a_{ {4}}a_{{2}}+a_{{3}}a_{{5}} \right) \lambda-{a_{{1}}}^{3}+2\,a_{{1}}a_{ {2}}a_{{4}}+a_{{1}}a_{{3}}a_{{5}}-{a_{{2}}}^{2}a_{{5}}-a_{{3}}{a_{{4}} }^{2}$$ It has $0$ as an eigenvalue with algebraic multiplicity $1$ if the constant coefficient is $0$ and the coefficient of $\lambda$ is nonzero, or multiplicity $2$ if both of these are $0$ but $a_1 \ne 0$.