Question: $A$ be $3\times3$ matrix with $\operatorname{trace}(A)=3$ and $\det(A)=2$. If $1$ is an eigenvalue of $A$ find eigenvalues for matrix $A^2-2I$?

In this question I got eigenvalues of $A=1,1+i,1-i$ . I was thinking to change $\lambda$ to $\lambda^2-2$ in characteristic equation of $A$ to get eigenvalues of required matrix. I m getting characteristic equation $x^6-9x^4+28x^2-30=0$. Let $x=\lambda$. Can anyone tell me why I m wrong? I m not getting results.

Answer: 
If $p(x)$ is a polynomial, the eigenvalues of $p(A)$ are the numbers $p(\lambda)$ for each eigenvalue $\lambda$ of $A$. What you are doing is instead taking the polynomial $f(p(x))$ where $f(x)$ is the characteristic polynomial of $A$, which is a totally different operation. The roots of $f(p(x))$ are numbers $a$ such that $p(a)$ is an eigenvalue of $A$, not numbers of the form $p(\lambda)$ such that $\lambda$ is an eigenvalue of $A$.
$A = P\pmatrix {1\\&1+i\\&&1-i}P^{-1}\\
A^2 = PDP^{-1}PDP^{-1} = PD^2P^{-1} = P\pmatrix {1\\&2i\\&&-2i}P^{-1}\\
A^2-2I = PD^2P^{-1} - 2PIP^{-1} = P(D^2-2I)P^{-1}\\
A^2-2I = P\pmatrix {-1\\&-2+2i\\&&-2-2i}P^{-1}$