Matrix and eigenvectors

Question

$\quad$The matrix $\mathbf A=\frac19\begin{bmatrix} 7 & -2 & 0 \\ -2 & 6 & 3 \\ 0 & 2 & 5 \\ \end{bmatrix}$ has eigenvalues $1$, $\frac23$ and $\frac13$n with the corresponding eigenvectors $ \mathbf v_1=\begin{bmatrix} -2 \\ 2 \\ 1 \\ \end{bmatrix}$, $\mathbf{v}_2=\begin{bmatrix} 2 \\ 1 \\ 2 \\ \end{bmatrix}$, $\mathbf v_3=\begin{bmatrix} 1 \\ 2 \\ -2 \\ \end{bmatrix}$.

$\text{(i)}$ Find the general solution of the equation $\mathbf x_{k+1}=\mathbf{Ax}_k$ if $\mathbf x_0=\begin{bmatrix} 1 \\ 11 \\ -2 \\ \end{bmatrix}$.

For this question, does it mean I should find out what $x_k$ is in terms of $A$? Can anyone help me please?

Answer 1

Here is a HINT: $$x_k = A^k x_0$$

Answer 2

You have

$$A=P^{-1}DP$$

With

$$P=\begin{bmatrix} -2&2&1\\ 2&1&2\\ 1&2&-2\\ \end{bmatrix}$$

and

$$D=\begin{bmatrix} 1&0&0\\ 0&2/3&0\\ 0&0&1/3\\ \end{bmatrix}$$

Then

$$x_n=A^nx_0= P^{-1}DP \; P^{-1}DP \dots P^{-1}DP =P^{-1}D^nP \,x_0$$

With

$$D^n=\begin{bmatrix} 1&0&0\\ 0&(2/3)^n&0\\ 0&0&(1/3)^n\\ \end{bmatrix}$$

After simplification of the matrix product, and noticing $P^2=9I_3$, thus $P^{-1}=\frac19P$,

$$x_n=\frac1{3^n}\begin{bmatrix} 2^{n+1}-4\cdot3^{n}+3\\ 4\cdot3^{n}+2^{n}+6\\ 2\cdot\left(3^{n}+2^{n}-3\right)\\ \end{bmatrix}$$