find an expression for $A^n$ for any positive integer N I have posted part of this question in another post which has gotten to long. I need help with the second part of the question which is finding an expression for $A^n$ for any positive integer $n$.
The previous question gave eigenvalues of $λ_1 =-\sqrt{2},  λ_2 =\sqrt{2} , λ_3 = 0$
and the corresponding eigenvectors are: 
$$ 
    v_1=\begin{pmatrix}
    1 \\ 1 \\ -\sqrt 2
    \end{pmatrix},\quad
    v_2=\begin{pmatrix}
    1 \\ 1 \\ \sqrt 2
    \end{pmatrix},\quad
    v_3=\begin{pmatrix}
    1 \\ -1 \\ 0
    \end{pmatrix}.
$$
The other question is here:
linear algebra eigenvalues/vectors and finding an expression for $A^n$
Which show the original question and previous working out. Any help would be much appreciated
 A: I see you have the eigenvectors and the eigenvalues of matrix $A$... So build a unitary matrix $U$ (in your case the matrix $U$ is orthogonal), with columns the eigenvectors of $A$. That is, $U=\begin{pmatrix}\ \frac{1}{2} &\frac{1}{2} &\frac{\sqrt{2}}{2}\\ \frac{1}{2} &\frac{1}{2} &\frac{-\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} &\frac{-\sqrt{2}}{2} &0 \end{pmatrix}.$ Then, $A^{n}=U^{*}\Lambda^{n}U=U^{T}\Lambda^{n}U$, as $U\in\mathbb{R^{n,n}}$. Matrix $\Lambda$ is a diagonal matrix with entries the eigenvalues of matrix $A$. So, $\Lambda^{n}=\begin{pmatrix} (-\sqrt{2})^n & 0 & 0\\ 0 & (\sqrt{2})^{n} & 0 \\ 0 & 0 & 0 \end{pmatrix}$. Note that the matrix $A$ is diagonalizable, but it is not nonsingular (invertible).

Right now i saw that $A=\begin{pmatrix} x & 0 & 1\\ 0 & x & 1 \\ 1 & 1 & x \end{pmatrix}$ and that you can write, $A=B+xI$. So $A^{n}=U^{T}\Lambda^{n}U$, but $\Lambda^{n}=\begin{pmatrix} (-\sqrt{2}+x)^n & 0 & 0\\ 0 & (\sqrt{2}+x)^{n} & 0 \\ 0 & 0 & x^{n} \end{pmatrix}$.
A: Equation $Ax = \lambda x$ gives you eigenvalue and eigenvectors. As you already found them out, try multiplying above equation by $A$ on both sides. It will tell you your answer.
As for representation you could check that $A$ is diagonalizable since all eigenvalues are different. 
And thus $A^n = P^{-1}D^nP$ where P is some invertible matrix and D is diagonal matrix with eigenvalues of $A$ on the diagonal.
