0
$\begingroup$

I've got the polynomial $P(z) = \Phi_0 - \Phi_1z $ defined by the following matrices of coefficients:

$$ \begin{eqnarray} \Phi_0 = \left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0.2 & 1 & 0 & 0 \\ -0.6 & -0.4 & 1 & 0 \\ 0 & 0.2 & 0.6 & 1 \end{array} \right] \quad \Phi_1 = \left[ \begin{array}{rrrr} 0 & 0 & 0.4 & 0.6 \\ 0 & 0 & 0 & 0.2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \,. \end{eqnarray} $$

I would like to know if there is some way to get the coefficients $\phi_{ij}$ knowing the roots of the polynomial.

The characteristic equation of the polynomial is given by:

$$ \begin{eqnarray} |\Phi_0 - \Phi_1 z| = \left| \begin{array}{rrrr} 1 & 0 & -0.4z & -0.6z \\ 0.2 & 1 & 0 & -0.2z \\ -0.6 & -0.4 & 1 & 0 \\ 0 & 0.2 & 0.6 & 1 \end{array} \right| = 0 \,, \end{eqnarray} $$

which yields the following second order equation:

$$ -0.0096 z^2 + 0.0432 z + 1 = 0 \,. $$

Solving this equation, the roots of the polynomial are found to be $-8.20$ and $12.70$. This shows how to get the roots when the coefficients are known. But what about the other way around? is it possible to get the coefficients $\phi_{ij}$ from the roots?

With a polynomial consisting of scalars rather than matrices, the coefficients can be obtained from the roots of the polynomial. It is discussed here. In that setting, it is relatively easy to implement the idea. For example, the roots of the polynomial $P(z) = 1 -0.5z + 0.2z^2$ are $1.25 \pm 1.85i$; then, upon the values of the roots, the coefficients can be obtained as follows:

# R code
roots <- polyroot(c(1, -0.5, 0.2))
coefs <- 1
for (z in roots)
  coefs <- c(coefs, 0) - c(0, coefs) / z
coefs[-1]
#[1] -0.5+0i  0.2+0i

Is there a similar procedure that returns the coefficients of the matrices $\Phi_0$ and $\Phi_1$ from the roots obtained above?

$\endgroup$
1
$\begingroup$

With a scalar polynomial of degree $n$, one has $n$ roots and $n$ coefficients. Hence it's not so shocking that one can obtain one from the other and vice versa. But in your 4-by-4 matrix setting, you've got 2 roots and 16 coefficients for each of your two matrices. Hence you cannot hope to obtain all information in the latter from the former. (The situation may improve somewhat depending on what assumptions you can make on the form of your matrices, but getting a unique answer will require that these constraints be quite strong.)

$\endgroup$
  • $\begingroup$ Actually I have fewer coefficients (the zeros and ones are fixed) but still it seems I am trapped by the issue that you mention. I don't think I could afford setting as many restrictions as necessary, so I will have to reconsider the whole thing. $\endgroup$ – javlacalle Jun 6 '16 at 11:22
  • $\begingroup$ I figured your matrices were more specialized than I assumed in my answer, but didn't want to jump to conclusions about what they 'should be.' But yes, you'd probably still end up with too many parameters free to get a unique solution. $\endgroup$ – Semiclassical Jun 6 '16 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.