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I have a recursion relation in the form of the following two equations:

$X_{t+1} = X_t + V_{t+1} \\ V_{t+1} = wV_t + cy(g-X_t)$

I want to write these two equations into a matrix form so that I can analyze using some Linear algebra techniques. the problem is, I cannot seem to write as a square matrix.


Here is my partial solution:

$X_{t+1} = X_t + wV_t + cy(g-X_t)\\ V_{t+1} = wV_t + cy(g-X_t)$

I will group all $X_t$ and $V_t$ together:

$X_{t+1} = (1-cy)X_t + wV_t + cyg \\ V_{t+1} = -cyX_t + wV_t + cyg$

If I set $g=0$, then this would come into a nice clean form as follows:

$X_{t+1} = (1-cy)X_t + wV_t \\ X_{t+1} = -cyX_t + wV_t$

So this can be written as follows:

$\begin{pmatrix} X_{t+1} \\ V_{t+1} \end{pmatrix} = \begin{pmatrix} 1-cy & w \\ -cy & w\end{pmatrix} \begin{pmatrix}X_t \\ V_t\end{pmatrix}$

The matrix therefore would be easy to diagonalize, etc. But we made a simplying case when $g=0$. Can I still make this form for the case when $g \neq 0$? How can I change the matrix?

Thanks

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1 Answer 1

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You are basically asking how to recast the nonhomogeneous matrix equation $$\begin{pmatrix} X_{t+1} \\ V_{t+1} \end{pmatrix} = \begin{pmatrix} 1-cy & w \\ -cy & w\end{pmatrix} \begin{pmatrix}X_t \\ V_t\end{pmatrix}+\begin{pmatrix}cyg \\ cyg\end{pmatrix} $$ into a homogeneous one.

To save writing, let us call your matrix M, the left hand side vector $\vec{Y_{t+1}}$, and the constant vector on the right hand side $\vec{C}$, so your equation now presents like $$ \vec{Y_{t+1}} =M \vec{Y_{t}} + \vec{C}~. $$

Now define the telescopic variable $$ \vec{Z_t}\equiv \vec{Y_{t+1}} -\vec{Y_{t}}= (M-I)\vec{Y_{t}}+\vec{C} ~, $$ so that $$\vec{Z_{t+1}}= \vec{Y_{t+2}} -\vec{Y_{t+1}}\\ =M \vec{Y_{t+1}} + \vec{C} - M \vec{Y_{t}} -\vec{C} = M \vec{Z_{t}} ~, $$ a homogeneous recursion, all right.

You may now proceed to study the eigenvalues of M, and so on and so forth.... to determine the orbit of your vector.

See WP.

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