Question: generalized eigenvectors are linearly independent proof?
Answer: A vector v∈V∖{0} is called a generalized eigenvector to an eigenvalue λ∈C of T iff there exists k>0 such that (λI−T)kv=0. From the generalized eigenspace decomposition it follows that generalized eigenvectors to different eigenvalues are linearly independent. hence a1=0, and a2=0.