Linear difference equations and kernels A linear difference equation with constant coefficients can be written as a polynomial in the shift and identity operators (denoted here by $S$ and $I$), and then factored.
$$0 = x_{n+2} - 5x_{n+1} + 6x_n = (S^2 - 5S + 6I)x_n = (S-3I)(S-2I)x_n$$
The solution space of the equation is the kernel of $S^2-5S+6I$. In practice, you find the kernel of $S-3I$ (being $x_n = c_03^n$), the kernel of $S-2I$ (being $x_n=c_12^n$) and you add the general forms of both solutions.
I have two questions concerning this method.


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*In general, the factors of the polynomial are of the form $(S-\lambda I)^m$ where $\lambda$ is real and $m$ is an integer (the multiplicity). If the method from above is to always work, the kernel of the composition $(S-\lambda I)^m(S-\mu I)^n$ should be equal to the direct sum of the individual kernels of each factor (given that $\lambda\neq\mu$). Why is this true?

*In the context of differential equations, a very similar method is used with the differential operator $D$ instead of the shift operator $S$. Again, the kernel of $(D-\lambda I)^m(D-\mu I)^n$ seems to be equal to the direct sum of the two individual kernels. Is the argument similar to the argument for my first question?


After some working, I managed to find a solution for this problem. In general, for any two commuting linear maps $A, B\colon V\to V$ on a vector space $V$ we have
$$\ker A + \ker B \subseteq \ker(AB).$$
This is because a vector in $\ker A$ gets mapped to zero under $AB = BA$ (and similarly for vectors in $\ker B$).
Adapting this to our context, we see that
$$\ker (S-\lambda I)^m + \ker (S-\mu I)^n \subseteq \ker\big((S-\lambda  I)^m(S-\mu I)^n\big).$$
Now each of the three kernels corresponds to a linear difference equation with constant coefficients of order $m$, $n$ and $m+n$. Since the solution space of such an equation has its dimension equal to the order of the equation, the dimensions in the relation above are equal and hence the vector spaces are equal.
