Let A be a given 4x4 matrix. Find matrices $F_1,F_2,F_3$ such that $A=c_1F_1+c_2F_2+c_3F_3$, $F_1+F_2+F_3=I$ and $F_iF_j=0$ when $i\not = j$ Let $A=\begin{bmatrix}0 & 1 &0 & 1\\1 & 0 & 1 & 0\\0 & 1 & 0 & 1\\1 &0 &1 &0\end{bmatrix}$
Find matrices $F_1,F_2,F_3$ such that $A=c_1F_1+c_2F_2+c_3F_3$, $F_1+F_2+F_3=I$ and $F_iF_j=0$ when $i\not = j$.
So this seems to be a direct application of this theorem:

My attempt:
Suppose $A$ is the representation of an operator $T$ in some basis (not yet defined).
The characteristic values for $A$ (and $T$) are:
 $c_1=0$, $c_2= 2$, $c_3=-2$.
Let $W_i$ be subspace of the vectors associated to $c_i$.
We then  have: 
$W_1=\langle(1,0,-1,0),(0,1,0,-1)\rangle$
$W_2=\langle(1,1,1,1)\rangle$
$W_3=\langle(-1,1,-1,1)\rangle$
So T is diagonalizable and we can apply the theorem.
So $E_1(a(1,0,-1,0)+b(0,1,0,-1)+c(1,1,1,1)+d(-1,1,-1,1))= a(1,0,-1,0)+b(0,1,0,-1)$ 
defines $E_1$.  It is analogous with $E_2$ and $E_3$.
We should have $T=c1E_1+c_2E_2+c_3E_3$ and so $[T] _B=c_1[E_1]_B+c_2[E_2]_B+c_2[E_3]_B$ for any basis B.
So now I choose B to be the basis $((1,0,-1,0),(0,1,0,-1), (1,1,1,1), (1,1,1,1))$ and define A to be the representation of T in that basis. Then 
$[E_1]_B=\begin{bmatrix}1 & 0 &0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & 0\\0 &0 &0 &0\end{bmatrix}$
$[E_2]_B=\begin{bmatrix}0 & 0 &0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 &0 &0 &0\end{bmatrix}$
$[E_3]_B=\begin{bmatrix}0 & 0 &0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 &0 &0 &1\end{bmatrix}$
To my sorrow, these matrices accomplish all the conditions except the first one. Where did I go wrong?
Note: I realized that its not necessary to compute basis for the $W_i’s$, since whatever basis I found  for each one, I will suppose that $A$ represents $T$ in that basis and so I will have for the $F_i's$ just ones and zeros depending on the dimension of the $W_i’s$.
 A: I didn't check your calculations, but assuming they are correct, you have found a basis of eigenvectors $\mathcal{B} = (v_1, \dots, v_4)$ for the matrix $A$ (or, equivalently, the operator $T = T_A \colon \mathbb{F}^4 \rightarrow \mathbb{F}^4$ defined using $A$ by left multiplication so $T_A(x) := Ax$). Since $\mathcal{B}$ is a basis of eigenvectors and $E_i$ are the corresponding projections, you indeed should have
$$ [T]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & -2 \end{pmatrix}, \\
[E_1]_{\mathcal{B}} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, [E_2]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, [E_1]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
and so $[T]_{\mathcal{B}} = c_1 [E_1]_{\mathcal{B}} + c_2 [E_2]_{\mathcal{B}} + c_3 [E_3]_{\mathcal{B}}$ and in particular $T = c_1 E_1 + c_2 E_2 + c_3 E_3$. Looking at the representing matrices with respect to the standard basis $\mathcal{C}$ of $\mathbb{F}^4$, we must also have
$$ A = [T]_{\mathcal{C}} = c_1 [E_1]_{\mathcal{C}} + c_2 [E_2]_{\mathcal{C}} + c_3 [E_3]_{\mathcal{C}} $$
and so the matrices you are looking for are
$$ F_i := [E_i]_{\mathcal{C}} = [\operatorname{id}]_{\mathcal{C}}^{\mathcal{B}} [E_i]_{\mathcal{B}} [\operatorname{id}]_{\mathcal{B}}^{\mathcal{C}} = P [E_i]_{\mathcal{B}} P^{-1}$$
where $P = [\operatorname{id}]_{\mathcal{C}}^{\mathcal{B}}$ is a change of basis matrix whose columns are simply the eigenvectors $v_1, \dots, v_4$ of $A$.
