Try to find some number if they exist, using eigenvalue

Question

Are there any value $a,b,c,d,e,f \in R$ such that

A=$\begin{bmatrix} 1& a& b \\ c& 1& d \\ e & f& 1 \end{bmatrix}$.

and $A^{2}=2A$? Try to find that value but not solving any system.

My opinion that is not exist, because using Cayley-Hamilton theorem, we can write $A(A-2I)=0$ so $\lambda1=0,\lambda2=2$, now $tr(A)=3$ so $\lambda3=1$, so $\sigma(A)=\{0,1,2\}$ know $\sigma(2A)=\{0,2,4\}$, $\sigma(A^2)=\{0,1,4\}$,but how $A^2=2A$ that mean that $\sigma(2A)=\sigma(A^2)$, but we can see that is not true, so it does not exist $a,b,c,d,e,f.$ What do you think?

Answer 1

You have the idea, $A(x)=cx$ implies that $c=0,2$ and $tr(A)$ is even, but $tr(A)=3$.