Let $A$ be the $4 \times 4$ matrix on $\mathbb{R}$
$A = \left[\begin{array}{cccc} 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ \end{array}\right ]$
find matrices $E^{'}_{1}, E^{'}_{2}$ and $E^{'}_{3}$, with $ E^{'}_{1} + E^{'}_{2} + E^{'}_{3} = I$ and $E^{'}_{i}E^{'}_{j} = 0$ if $ i \neq j$.
My try:
i let $A$ define a operator $T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}$. By calculating the characteristic polynomial of $T$ we notice that $T$ is diagonalizable and his eigenvalues are $0, 2$ and $-2$. So i compute the Kernel of the respectives characteristic spaces and i got $dim (Ker T) = 2$, $dim (Ker T-2I) = 1$, and $dim (Ker T+2I )= 1$, then i define three operators with $E_{1}(\mathbb{R}^{4}) = Ker T$, $E_{2}(\mathbb{R}^{4}) = Ker (T-2I)$ and $E_{3}(\mathbb{R}^{4}) = Ker (T+2I)$, so by my calculating the respectives matrices are
$E_{1}^{'} = \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{array}\right ]$ $E_{2}^{'} = \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{array}\right ]$ and $E_{3}^{'} = \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{array}\right ]$ so
$c_{i}$ are the eigenvalues of $T$ and $A = 0E^{'}_{1} + 2E^{'}_{2} + (-2)E^{'}_{3}$, but that matrix $A$ its the diagonal form of $A$ how are the correct way to solve a questions like that?