I have a cubic equation, of which the roots are the eigenvalues of a 3×3 matrix. The coefficients depend on a parameter (z), and I want to prove that one of the eigenvalues ($\lambda_2$, the second one in magnitude) is locally increasing in $z$ (around $z=0$).
The characteristic polynomial has three distinct real, positive roots. The polynomial for $z \neq 0$ has the usual form:
$-\lambda^3 + b(z)\lambda^2 + c(z)\lambda + d(z) = 0$
where $b$, $c$ and $d$ are all linearly functions of $z$.
I can analytically derive the roots for $z=0$. In that case the characteristic polynomial is:
$(m-\lambda)( d_2\lambda^2 + d_1\lambda + d_0)$
where I can solve the second part with the quadratic formula to get an analytical expression. But I want to know how the roots change with z.
I used this formula (see below) to obtain an analytical expression for the cubic roots. However I can not calculate the derivative of the roots with respects to z, because the quadratic root Q gives an imaginary number. This prevents me from differentiating the root with respect to z. From what I understand it is always the case that you need imaginary numbers to get an algebraic expression for the cubic roots, if you have three real roots.
Anybody a suggestion how I can calculate the marginal change in one particular eigenvalue ($\lambda_2$), ie. $\frac{\partial \lambda_2}{\partial z}$ around $z=0$? As already mentioned, I do know the exact roots when $z=0$.
To be clear, I am not interested in a numerical solution. I want an expression in terms of the parameters.
Formula for cubic roots:
$a(z)x^3 + b(z)x^2 + c(z)x + d(z) = 0$
the roots are then (omitting that the coefficients depend on z):
$x = p + \sqrt[3]{q-Q} + \sqrt[3]{q+Q}$
with $p=-b/(3a)$, $q=p^3 – (pc+d)/2a$, $r = c/(3a)$, and $Q=\sqrt{(q^2+(r-p^2)^3)}$
Best Answer
How does the imaginary number prevents you from differentiating, though the complex number always cancels out Maybe you should check the trigonometric cubic formula
Speaking of your topic that's interesting, why should complex number appear when we don't need them?
Looking at the formula you posted $x = p + \sqrt[3]{q+Q} + \sqrt[3]{q-Q}$, the cube root part are conjugate and we know in normal cases, conjugates can be simplified
$(q+Qi)(q-Qi) = q^2+Q^2$ $(q+Qi)+(q-Qi) = 2q$ $(q+Qi)^2+(q-Qi)^2 = 2q^2-2Q^2$ $(q+Qi)^3+(q-Qi)^3 = 2q^3-6qQ^2$ $\sqrt{q+Qi}+\sqrt{q-Qi} = \sqrt{2q±2\sqrt{q^2+Q^2}}$ $\sqrt[3]{q+Qi}+\sqrt[3]{q-Qi} = ?$
the cubic formula can't be simplified, because it's already in its simplest case
check this cubic equation $x^3-6x-6=0$, $x = \sqrt[3]{2}+\sqrt[3]{4}$
Just like the quadratic formula makes use of the square root of unity, $\sqrt{1} = \pm 1$
So the truth behind this problem is that the cubic formula makes use of the cube root of unity $\sqrt[3]{1} = 1$, $-\frac{1}{2}+\frac{\sqrt{-3}}{2}$ and $-\frac{1}{2}-\frac{\sqrt{-3}}{2}$ meaning that it would always have complex number
In general every polynomial of degree $n$ makes use of the nth-root of unity in its formula