yes. the sum total of the dimensions of the eigenspaces equals the dimension of the domain of A. therefore, we have an eigenbasis for the domain of A. if a basis for the eigenspace corresponding to λ_{1}is {v_{1}, v_{2}, v_{3}} and a basis for the eigenspace corresponding to λ_{2}is {v_{4}, v_{5}}, then the matrix P whose columns are the eigenbasis for V satisfies:

AP = PD, where D = diag{λ_{1}, λ_{1}, λ_{1}, λ_{2}, λ_{2}} so that:

P^{-1}AP = D, a diagonal matrix.

the invertibility of P is guaranteed by the linear independence of the eigenvectors (the only vector E_{λ1}and E_{λ2}have in common is the zero-vector:

if Av = λ_{1}v = λ_{2}v, then:

(λ_{1}- λ_{2})v = 0, since λ_{1}≠ λ_{2}, we must have v = 0).