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Say you have a reucrrence $x_{n+1} = 3x_n+2$. Though, it is a inhomogenous, it can be represented by a linear system

$$\begin{bmatrix} x_{n+1}\\1\end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 0 & 1\end{bmatrix}\, \begin{bmatrix} x_n\\1\end{bmatrix}.$$

In case $X_n=\begin{bmatrix} x_{n}\\1\end{bmatrix}$ is an eigenvector,

$$X_{n+1} = \begin{bmatrix} 3 & 2 \\ 0 & 1\end{bmatrix}\,X_n = \begin{bmatrix} 3 & 2 \\ 0 & 1\end{bmatrix}\, \begin{bmatrix} x_n\\1\end{bmatrix} = s\,\begin{bmatrix} x_{n}\\1\end{bmatrix} = sX_n.$$

Applying eigenvector is identical to scaling it by eigenvalue, $s$.

I have seen somewhere that I can use powers of $s$ to represent $x_n$: $\begin{bmatrix} x_{n+1}\\x_n\end{bmatrix} = \begin{bmatrix} s^{n+1}\\s^n\end{bmatrix}$. Since s is also an eigenvalue, this may lead to some confusion. Neverless, I think that using the polinomials of eigenvalue symbol is worth and popular in Laplace domain, where you have powers of $s$ for time shifting. You multiply you vector by $s$ and get the next state. This blurs the border between applying the operator and scaling with eigenvalue. Such abuse of notation is also advantageous because we clearly see why $X_{n+1} = sX_n$: $X_2 = \begin{bmatrix} s^2\\s\end{bmatrix} = s\,\begin{bmatrix} s\\1\end{bmatrix} = s X_1.$ It was not that visible in case of $X_{n} = \begin{bmatrix} x_n\\x_{n-1}\end{bmatrix}.$

So, recurrence is $s^2 = 3s + 2$ and, in case $s$ is eigenvalue, matrix form is

$$\begin{bmatrix} s^2\\s\end{bmatrix} = s\,\begin{bmatrix} s\\1\end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 0 & 1\end{bmatrix}\, \begin{bmatrix} s\\1\end{bmatrix}.$$

Normally, vector components are indexed. Every next component is index+1 the previous one. In case indexes are replaced with powers, we have the components related by s: all components of our eigenvectors are powers of eigenvalue s! This should work for every diagonalizable matrix. I just came up with concrete example to examine some concrete eigenvectors.

You may spot the eigenvalues. The system has a feedback of power 3, resulting in the first eigenvalue $s=3$ (it is coupled with eigenvector $X=\begin{bmatrix} 1\\0\end{bmatrix}$). The second eigenvalue, $s= 1$, keeps the contribution 2 constant (it is coupled with eigenvector $\begin{bmatrix} -1 \\ 1\end{bmatrix}$). The problem is that I see that the components of these vectors, $\begin{bmatrix} 1\\0\end{bmatrix}$ and $\begin{bmatrix} -1 \\ 1\end{bmatrix}$ are not the powers of $s$. They do not have the form $\begin{bmatrix} s^{n+1} \\ s^{n}\end{bmatrix}$! Why they should not?

Val
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    Shouldn't that $1$ in the bottom right entry of the matrix be in the bottom left entry, and the $0$ be in the bottom right entry? – Alex Wertheim Jul 29 '13 at 21:15
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    Yes, your matrix should be $\begin{bmatrix}3&2\1&0\end{bmatrix}$ – Thomas Andrews Jul 29 '13 at 21:23
  • Thanks for correcting but the system stems from the $\begin{bmatrix} x_2\1\end{bmatrix} = \begin{bmatrix} 3 & 2 \ 0 & 1\end{bmatrix}, \begin{bmatrix} x_1\1\end{bmatrix}$, which describes recurrence $x_2=3x_1+2$. One must be in the right. I just presented it as polynomial with $x_n=s^n$. Don't forget that $s$ just stands for eigenvalue (or shift operator, as it is often called) and it works when we plug the eigenvectors. – Val Jul 29 '13 at 21:36

1 Answers1

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A linear homogeneous recurrence with constant coefficients, such as $$ x_n=c_1x_{n-1}+\ldots+c_kx_{n-k}, $$ can be expressed as a matrix equation of the form $X_n=AX_{n-1},$ namely $$ \begin{bmatrix} x_n\\ x_{n-1}\\ x_{n-2}\\ \vdots\\ x_{n-k+1}\end{bmatrix} = \begin{bmatrix} c_1 & c_2 & \ldots & c_{k-1 }& c_k\\ 1 & 0 & \ldots & 0 & 0\\ 0 & 1 & \ldots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \ldots & 1 & 0 \end{bmatrix}\, \begin{bmatrix} x_{n-1}\\ x_{n-2}\\ \vdots\\ x_{n-k+1}\\ x_{n-k} \end{bmatrix}. $$ The $k\times k$ matrix $A$ in this equation has an eigenvector $\begin{bmatrix}s^{k-1} & s^{k-2} & \ldots & 1\end{bmatrix}^T$ with eigenvalue $s$ for every distinct $s$ satisfying $s^k=c_1s^{k-1}+c_2s^{k-2}+\ldots+c_k.$ The recurrence therefore has a solution $\left[x_0,x_1,x_2,\ldots\right]=\left[1,s,s^2,\ldots\right]$ for each such $s.$ If the eigenvalues are all distinct, then all solutions are linear combinations of such solutions.

An inhomogeneous recurrence, as in your problem, is not expected to have solutions of this form. Your statements, $$ X_{n+1}=sX_n,\qquad X_2=\begin{bmatrix}s^2\\ s\end{bmatrix}, $$ do not hold. Indeed, the second component of your vectors $X_n$ is fixed at $1$ for all $n,$ so you cannot have solutions whose components are powers of $s.$ (Except possibly if $s=1,$ but this leads to $X_n=\begin{bmatrix}1 & 1\end{bmatrix}^T$ for all $n,$ which does not solve your recurrence.

There is no mystery here: the matrix $A$ above has $1$s below the diagonal, whereas your matrix, $$ \begin{bmatrix} 3 & 2\\ 0 & 1 \end{bmatrix}, $$ has $1$ on the diagonal. The form of solution that works for $A$ has no reason to work for matrices of your form.

Will Orrick
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