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When $A$ is diagonalisable, $\vec{x}_{k+1}=A\vec{x}_k$ implies that $\vec{x}_k = c_1\lambda_1^k\vec{v}_1 +...+c_n\lambda_n^k\vec{v}_n$ because an eigenbasis exists and any $\vec{x}$ can be decomposed into a linear combination of $A$'s eigenvectors.

But when $A$ is not diagonalisable, is there a related general formula for (or a general technique to find) $\vec{x}_k$ in terms of $A$'s eigenvalues and eigenvectors?

I ask this because I have just learnt that for the differential system $\dot{\vec{x}}=A\vec{x}$, a fundamental solution set always exists so that even when $A$ is defective, $\vec{x}$ can still be expressed in terms of $A$'s eigenvectors and eigenvalues.

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It works pretty much the same way as for differential equations. Let $P^{-1}AP=J$ where $J$ is in Jordan form. Then $x_k=A^kx_0=PJ^kP^{-1}x_0$ and you just need to understand powers of a matrix in Jordan form, and the relation of $P$ to the eigenvectors and generalized eigenvectors.

EDIT: Here's the simplest example. Say $A=\pmatrix{\lambda&1\cr0&\lambda\cr}$ Then $A^n=\pmatrix{\lambda^n&n\lambda^{n-1}\cr0&\lambda^n\cr}$ Let $v_1$ be the eigenvector $(1,0)$, let $v_2$ be the generalized eigenvector $(0,1)$. Then $A^nv_1=\lambda^nv_1\quad A^nv_2=n\lambda^{n-1}v_1+\lambda^nv_2$ so $A^n(c_1v_1+c_2v_2)=(c_1\lambda^n+c_2n\lambda^{n-1})v_1+c_2\lambda^nv_2$

But do check my arithmetic.

MORE EDIT: We can rewrite that last expression as $x_n=c_1\lambda^nv_1+c_2\lambda^n((n/\lambda)v_1+v_2)$ which looks a little more like the form you've given for the analogous difference equation.

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    I'm very lost. Will you express $\mathbf x_k$ in terms of the eigenvecs and eigenvals? (For a differential system containing 2 equations and where $\lambda$ has algebraic multiplicity 2 and geometric multiplicity 1, we have $\mathbf x=c_1e^{\lambda t}\mathbf v+c_2 e^{\lambda t} (t\mathbf v +\mathbf w)$. Is there a related equation for the analogous difference system?)2012-08-26