I can't proceed with linear algebra for eigenvector and eigenvalue.

Question

Q. Find the solution of the difference equations $x_n+_1$ = A$x_n$ with initial condition $x_0$ at time n = 1004 when $A$ is\begin{bmatrix}1&1\\4&-2\end{bmatrix} and $x_0$ is \begin{bmatrix}1\\2\end{bmatrix}

I find eigenvectors and eigenvalues $v_1$ = \begin{bmatrix}-1\\4\end{bmatrix} $v_2$ = \begin{bmatrix}1\\4\end{bmatrix}

but i can't proceed more.

Answer 1

You made a mistake with your eigenvectors. You should instead have $$ \lambda_1 = -3,\ v_1 = (-1,4) \qquad \lambda_2 = 2, \ v_2 = (1,1). $$ Now, solve the system of equations $$ c_1 v_1 + c_2 v_2 = x_0. $$ Once you have the numbers $c_1,c_2$, we have $$ A^n x_0 = A^n(c_1 v_1 + c_2 v_2) = c_1 A^n v_1 + c_2 A^n v_2. $$