True or false matrix question $A^2=2A$ Is it true that $a, b, c, d, e, f \in R$ exist so that the following is possible?
$$A=\begin{bmatrix}
1 && a && b \\
c && 1 && d \\
e && f && 1
\end{bmatrix}
$$
$A^2=2A$
I am not allowed to solve any kind of system with $a, b, c, d, e, f$ so I don't really know how to even approach this problem.
 A: The fact that $A^2=2A$ implies $A^2-2A=0$ is a matrix polynomial equaling zero.
As any matrix has a minimal polynomial and any matrix polynomial equaling zero is a multiple of the minimal polynomial, we learn that the minimal polynomial is one of:
$$x, x-2,~\text{or}~x^2-2x$$
This implies further that the eigenvalues of the matrix are some combination of $0$'s and $2$'s.
Note now that the trace of your matrix is equal to $3$ which could not be the sum of the eigenvalues (possible sums being $0+0+0,0+0+2,0+2+2,2+2+2$), a contradiction.

Related: If $A^2=2A$, then $A$ is diagonalizable.
A: We have 
\begin{eqnarray*}
A=\begin{bmatrix}
1 && a && b \\
c && 1 && d \\
e && f && 1
\end{bmatrix}\end{eqnarray*}
Now calculate
\begin{eqnarray*}
A^2 &=&\begin{bmatrix}
1 +ac+be && 2a+bf && 2b+ad \\
2c+de && ac+1+df && 2d+cb \\
2e+cf && 2f+ae && be+df+1
\end{bmatrix} \\
2A&=&\begin{bmatrix}
2 && 2a && 2b \\
2c && 2 && 2d \\
2e && 2f && 2
\end{bmatrix}
\end{eqnarray*}
Consider the three equations on the leading diagonal
\begin{eqnarray*}
ac+be=1 \\
ac+df=1 \\
df+be=1 
\end{eqnarray*}
So $ac=be=df=\frac{1}{2}$. Now consider the $(1,2)$ entry $2a+bf=2a$, So $bf=0$ which contradicts $be=\frac{1}{2}$ or $df=\frac{1}{2}$.
