## Wednesday, September 30, 2009

### Solving an n-order polynomial by converting into a linear matrix reverts to the polinomial

Here is an attempt to find the polinomial roots by converting it into a system of the same order, n, by following the typical way to represent an n-order differential equation in matrix form of n first order equations. Polinomial

sn + a1 sn-1 + .. + an = 0.

I substitute x1= s and xn-1= xn /s so that

sxn-1 = xn= -(a1 xn-1 + a2 xn-2 + .. + an x0),

which may be presented in vector-matrix form:

$$\begin{bmatrix} x_n \\ x_{n-1} \\ \vdots \\ x_1 \end{bmatrix} = s\,\begin{bmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_0 \end{bmatrix} = sX =$$ $$= \begin{bmatrix} -a_1 & -a_2 & \cdots & -a_n \\ 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_0 \end{bmatrix} =AX$$ wherein X stands for the column vector of accumulator variables and A is the recurrence operator.

Now, we need to find s. It turns out to be an eigenvalue. Finding an eigenvalue implies solving the characteristic plynomial. So we are back at the start. My trick to find s by solving a system of linear equations had not succeeded.

The idea of perfect match between the recurrence and its characteristic polynomial was elaborated in Characteristic polynomial - 3