While studying linear algebra, I became interested in the characteristic polynomial. I identified several types of matrices that interest me and made a system of equations for them.
I am trying to solve a nonlinear system of equations for the variables $b_i$, where each equation is obtained as a coefficient of $s$. For a matrix of dimension 3, I was able to find a solution on my own, but for higher orders, nothing works.
Is there an easy way to solve such systems?
$$ M_4(b_1,b_2,b_3,b_4)=\left( \begin{array}{ccccc} s-\lambda _1 & -b_2 & -b_3 & -b_4 & -1 \\ b_1 & s-\lambda _2 & -b_3 & -b_4 & -1 \\ b_1 & b_2 & s-\lambda _3 & -b_4 & -1 \\ b_1 & b_2 & b_3 & s-\lambda _4 & -1 \\ b_1 & b_2 & b_3 & b_4 & 0 \\ \end{array} \right) $$
$$ Det(M_4(b_1,b_2,b_3,b_4))=-b_2 b_3 b_4 \lambda _1-b_1 b_3 b_4 \lambda _2-b_1 b_2 b_4 \lambda _3-b_4 \lambda _1 \lambda _2 \lambda _3-b_1 b_2 b_3 \lambda _4-b_3 \lambda _1 \lambda _2 \lambda _4-b_2 \lambda _1 \lambda _3 \lambda _4-b_1 \lambda _2 \lambda _3 \lambda _4+s^2 \left(-b_2 \lambda _1-b_3 \lambda _1-b_4 \lambda _1-b_1 \lambda _2-b_3 \lambda _2-b_4 \lambda _2-b_1 \lambda _3-b_2 \lambda _3-b_4 \lambda _3-b_1 \lambda _4-b_2 \lambda _4-b_3 \lambda _4\right)+\left(b_1+b_2+b_3+b_4\right) s^3+s \left(b_3 \lambda _1 \lambda _2+b_3 \lambda _1 \lambda _4+b_3 \lambda _2 \lambda _4+b_4 \lambda _1 \lambda _2+b_2 \lambda _1 \lambda _3+b_4 \lambda _1 \lambda _3+b_1 \lambda _2 \lambda _3+b_4 \lambda _2 \lambda _3+b_2 \lambda _1 \lambda _4+b_1 \lambda _2 \lambda _4+b_1 \lambda _3 \lambda _4+b_2 \lambda _3 \lambda _4+b_1 b_2 b_3+b_1 b_4 b_3+b_2 b_4 b_3+b_1 b_2 b_4\right) $$
$$ \begin{cases} b_1+b_2+b_3+b_4=0 \\ -b_2 \lambda _1-b_3 \lambda _1-b_4 \lambda _1-b_1 \lambda _2-b_3 \lambda _2-b_4 \lambda _2-b_1 \lambda _3-b_2 \lambda _3-b_4 \lambda _3-b_1 \lambda _4-b_2 \lambda _4-b_3 \lambda _4=0 \\ b_3 \lambda _1 \lambda _2+b_3 \lambda _1 \lambda _4+b_3 \lambda _2 \lambda _4+b_4 \lambda _1 \lambda _2+b_2 \lambda _1 \lambda _3+b_4 \lambda _1 \lambda _3+b_1 \lambda _2 \lambda _3+b_4 \lambda _2 \lambda _3+b_2 \lambda _1 \lambda _4+b_1 \lambda _2 \lambda _4+b_1 \lambda _3 \lambda _4+b_2 \lambda _3 \lambda _4+b_1 b_2 b_3+b_1 b_4 b_3+b_2 b_4 b_3+b_1 b_2 b_4=0 \\ b_2 b_3 b_4 \lambda _1-b_4 \lambda _2 \lambda _3 \lambda _1-b_3 \lambda _2 \lambda _4 \lambda _1-b_2 \lambda _3 \lambda _4 \lambda _1-b_1 b_3 b_4 \lambda _2-b_1 b_2 b_4 \lambda _3-b_1 b_2 b_3 \lambda _4-b_1 \lambda _2 \lambda _3 \lambda _4=0 \end{cases} $$
I tried to compute with Wolfram Mathematica and got the following solution:
