It's a good thing you told me what book you were using in that other thread, because I'm not familiar with the terminology here. However, I looked your book up on Google Books, so we're good to go now!
I've heard spectral decomposition used sometimes as a synonym for diagaonlization, and that's essentially what we're doing when we look for these projectors.
Let [itex]A[/itex] be an nxn matrix. Suppose it is similar to a diagaonl matrix:
[tex]P^{-1} AP = \Lambda.[/tex]
The diagonal matrix [itex]\Lambda[/itex] contains the eigenvalues, as usual. Then
[tex]A = P\Lambda P^{-1} = P(\lambda_1 D_1 + \ldots +\lambda_nD_n)P^{-1} = \lambda_1 PD_1P^{-1} + \ldots + \lambda_n PD_nP^{-1},[/tex]
where the [itex]\lambda[/itex]'s are the eigenvalues (not necessarily unique) and the [itex]D_i[/itex] matrices are a zero matrix with a 1 in the [itex]i,i[/itex] position. Then, define
[tex]E_i = PD_iP^{-1}[/tex]
so that
[tex]A= \lambda_1 E_1 + \ldots + \lambda_nE_n.[/tex]
The right-hand side of the equation is termed the spectral decomposition of A and the E's are the projectors.
I see the Lagrange polynomials that you are referring to. They are in the proof of Theorem 7.11, right? I don't think we need to use those, we can just do this directly.
We first need to diagonalize our matrix A. You will NOT get 3 unique eigenvalues, but that's OK. A full set of unique eigenvalues is only a sufficient condition for diagonalization, not necessary. Once you diagonalize A, follow the formula I wrote above, [itex]E_i=PD_iP^{-1}[/itex], to get your ith projector. At the end, write down the decomposition using these matrices and see if you recover A.
EDIT: BTW, they are called projectors because they are idempotent, or [itex]E^2=E[/itex]. Any matrix that fixes its range is a projection/projector.