Diagonalizable Matrix Orthogonal Projections

Suppose A is diagonalizable with distinct eigenvalues...

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...cancels with the denominator.
Let v be an eigenvector corresponding to the eigenvalue aj, so Av=aj*v. then for i different from j we have (t-ai)Av=Av-ai*v=(aj-ai)v so when we compute the product from P_j we will get all the product (aj-ai) which is just c_j so:
P_j(A)(v)=c_j*v/c_j=v which shows that any vector v in the ...