# Proof for the solution of homogenous difference equation

Suppose the output $$y_h[n]$$ of a linear, time-invariant system is described by by the following equation for input $$x[n]=0$$,

$$\sum_{k=0}^{N}a_k y_h[n-k] = 0$$

My book states the $$y_h[n]$$ is in fact a member of a family of solution of the form:

$$y_h[n]=\sum_{m=1}^{N}A_mz_m^n$$

where the complex numbers $$z$$ are the zeros of $$A(z)=\sum_{k=0}^Na_kz^{-k}$$.

What is the proof that $$y_h$$ is in fact of the form given?

A simple substitution would result in:

$$\sum_{k=0}^N (a_k \sum_{m=1}^NA_mz_m^n) = 0$$

but I don't see to proof showing a set $$\{z_1,z_2,...,z_N\}$$ and $$\{A_1, A_2,..., A_N\}$$ indeed exist for all n. Sorry if this seems like a poor question.

• Do you have a definition for $A_m$? Mar 26 '20 at 16:51
• The first equation doesn't seem to me like a good description of the output. For example, you could make $a_k=0$ and then the equation is true, but it doesn't tell you anything about $y_h$. Also: the output of an LTI system to input $x[n]=0$ must be $y[n]=0$.
– MBaz
Mar 26 '20 at 16:59
• @zabop $A_m$ are the coefficients for output formula. Mar 26 '20 at 21:01
• @MBaz You're correct that the output of an LTI system for $x[n]=0$ must be $y[n]=0$. I think that right side of the difference equation $\sum_{m=0}^Mb_kx[n-m]$ disappears when $x[n]=0$ which allows you to solve for $y_h$ easily. Mar 26 '20 at 21:08
• This question is related to pg. 40 of Alan V. Oppenheim's book "Discrete-Time Signal Processing". There he described how to solve difference equations. Mar 26 '20 at 21:09

\begin{align}\sum_{k=0}^{N}a_k\sum_{m=1}^NA_mz_m^{n-k}&=\sum_{m=1}^NA_mz_m^n\sum_{k=0}^Na_kz_m^{-k}\end{align}\tag{1}
If $$z_m$$ are zeros of the polynomial
$$A(z)=\sum_{k=0}^Na_kz^{-k}\tag{2}$$
then Eq. $$(1)$$ equals zero, as required for the homogenous solution.