Prove that in any power of the matrix $$\begin{bmatrix} 2&1&1\\0&2&1\\1&1&1\end{bmatrix}$$ two entries on the main diagonal will be the same.
I was able to prove this by calculating the eigenvectors and eigenvalues, then constructing a diagonalisable matrix, resulting in me proving that the first and second elements on the main diagonal are always the same.
However, since the eigenvalues have many decimal places, I had to shorten them and include an ellipsis, and so I was told that even though I've correctly proved it for 5dp, it may not be true for 100dp for example.
Any other ideas are appreciated.
The claim is that the $(1,1)$ and $(2,2)$ entries are equal. Note that this is a $3 \times 3$ matrix, so its characteristic polynomial is a cubic $P(z)$, and it satisfies that characteristic polynomial. If your matrix is $A$, this means $A^m$ is a certain linear combination of $A^{m-1}, A^{m-2}, A^{m-3}$ for all $m \ge 3$. Once you see that the claim is true for $I$, $A$ and $A^2$, it is true by induction for all positive integers $m$.