Solve system in $a,b,c$

Question

What's the best way to solve the system of equations ?: $$ \left\{\begin{array}{rcccccr} a & + & b & + & c & = & 1 \\ a^{2} & + & b^{2} & + & c^{2} & = & 7 \\ a^{3} & + & b^{3} & + & c^{3} & = & 13 \end{array}\right. $$ If I square the first one I get $2\left(\,ab + bc + ac\,\right)=-6$, and if I multiply the second by $a + b + c$ I get some other expression that equals $-6$ but I'm not sure how to proceed.

Answer 1

By using Newton's identities, we can find that $$s_2:=ab+bc+ca=\frac{1}{2}(a+b+c)^2-\frac{1}{2}(a^2+b^2+c^2)=-3$$ and $$s_3:=abc=\frac{1}{3}(a^3+b^3+c^3)-\frac{1}{3}(a+b+c)^3+(ab+bc+ca)(a+b+c)=1.$$ Then we obtain $(a,b,c)$ by solving the polynomial equation $$x^3-s_1x^2+s_2x-s_3=x^3-x^2-3x-1=(x+1)(x^2-2x-1)\\=(x+1)(x-(1+\sqrt{2}))(x-(1-\sqrt{2}))=0$$ where $s_1=a+b+c$.

Answer 2

A result which may help if you know about matrices and linear algebra.

A diagonalizable matrix $A$ with eigenvalues $a,b,c$ will have \begin{align}\text{Tr}(A) &= 1\\ \text{Tr}(A^2) &= 7\\ \text{Tr}(A^3) &= 13\end{align}

Now you can use your favourite linear algebraic numerical method to solve for it.

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EDIT One linearization you can try is to start with $A = $ zero matrix, minimize

$$\min_{V} \{\|Tr(A+V) - 1\|_2^2 + \|Tr((A+V)A) - 7\|_2^2 + \|Tr((A+V)A^2) - 13\|_2^2 + \\\|V\|_2^2 + R(V+D)\}$$

followed by $A = A+V$ and iterating, solving for a new $V$ for this new $A$ and so on.

where $R(V+D)$ is a regularizer punishing non-diagonal elements of $V+D$ to differ from zero.

You can implement it with Kronecker products:

1. The "Tr" becomes a scalar product with a vector of 0s and 1s on the vectorization of matrices.
2. Multiplication with $A$ or $A^2$ can be represented like this.

Answer 3

Here is a graphical solution:

We use the first equation to eliminate $z = 1 - x -y$.

Then the second equation turns into an ellipse (red curve) and the third into some algebraic curve (green curve).

We can find six intersections and thus solutions of the system.