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Can anyone provide the forms of the solutions for the homogeneous part and particular solutions for a non-homogeneous system of two linear autonomous difference equations $\mathbf{X_{t+1}}=\mathbf{AX_t}+\mathbf{b}$, where $\mathbf{A}$ is a 2x2 matrix of parameters and $\mathbf{b}$ is a 2x1 vector of constants; when $\mathbf{A}$ has two (repeated) eigenvalues $\lambda_1=\lambda_2=1$. Thank you.

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A non-homogeneous system with a unit root has no steady state! – user58641 Mar 21 '13 at 21:40
Stability, that's what I mean. The general formula should not change. – user60610 Mar 22 '13 at 17:48

It is easy to check by induction that the solution is: $x_{t} = A^t x_0+ \sum_{k=0}^{t-1} A^k b$.

(If $A$ was invertible we would have $I+\cdots +A^{t-1} = (I-A)^{-1} (I-A^t)$, and the solution would 'simplify' to $x_{t} = A^t x_0+ (I-A)^{-1} (I-A^t)b$. However, since the eigenvalues are $1$, this does not apply here.)

Hence the homogenous part is $t \mapsto A^t x_0$ and the particular part is $ t \mapsto \sum_{k=0}^{t-1} A^k b$.

If $A$ is diagonalizable, then $A = U \Lambda U^{-1} = U U^{-1} = I$, so we must have $A^t = I$.

If $A$ is not diagonalizable, then $A = U \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} U^{-1}$, and $A^t = U \begin{bmatrix} 1 & t \\ 0 & 1\end{bmatrix} U^{-1} = I + t U \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} U^{-1} = I + t B$.

Therefore the general form of solution is $x_{t} = (I+tB)x_0+ (tI+\frac{t(t-1)}{2}B) b$, where the value of $B$ depends on the specific form of $A$.

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Yes of course, but how does this translate to the compact solution expressed with Eigenvalues and Eigenvectors? – user58641 Mar 21 '13 at 21:39
Well, this depends on the form of $A$, $A$ may not have a basis of eigenvectors. See above for elaboration. – copper.hat Mar 21 '13 at 22:09
Very helpful thank you. However, this is pushing my knowledge of matrix algebra a little hard - sorry! Now, in the non-diagonalizable case, isn't $\mathbf{B}$ itself a function of time and therefore needing to be computed with each $t$? – user58641 Mar 21 '13 at 22:35
I guess my real question there is really how do I compute $\mathbf{B}$ – user58641 Mar 21 '13 at 22:56
No, $B$ is given by $U \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} U^{-1}$ which is time independent. – copper.hat Mar 21 '13 at 23:01

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