TITLE: generating matrix for a recurrence relation
QUESTION [6 upvotes]: for the recurrence f(n)=a*f(n-1)+b*f(n-2)+c*f(n-3)+d*f(n-4) , how can one get the generating matrix so that it can be solved by matrix exponentiation?
For f(n)=a*f(n-1)+b*f(n-2)+c*f(n-3) the corresponding generating matrix is: 
| a  b  c |   |  f(n)  |   | f(n+1) |
| 1  0  0 | x | f(n-1) | = |  f(n)  |
| 0  1  0 |   | f(n-2) |   | f(n-1) |

so how to get the same for required recurrence?
Also what should be the procedure for any recurrence which may be of the form :
$f(n)=a_1f(n-1)+a_2f(n-2)+a_3f(n-3)+..+a_kf(n-k)$ ?
Thanks.

REPLY [10 votes]: You can easily verify the answer below by applying the matrix multiplication.
State transition matrix for the given recurrence-relation is
$$
\begin{bmatrix}
f(n) \\
f(n-1) \\
f(n-2) \\
\vdots \\
f(n-(k-1))
\end{bmatrix} = \begin{bmatrix}
       a_1 & a_2 & a_3 & \dots & a_{k-1} & a_k           \\[0.3em]
       1 & 0 & 0 & \dots & 0 & 0 \\[0.3em]
       0 & 1 & 0 & \dots & 0 & 0 \\[0.3em]
       0 & 0 & 1 & \dots & 0 & 0 \\[0.3em]
       \vdots & & & \ddots & 0 & 0 \\[0.3em]
       0 & 0 & 0 & \dots & 1 & 0 \\[0.3em]


     \end{bmatrix} \begin{bmatrix}
f(n-1) \\
f(n-2) \\
f(n-3) \\
\vdots \\
f(n-k)
\end{bmatrix} $$