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**Ackbeet** In finite-dimensional vector spaces, which is what you have, linear operators ARE matrices (at least, there's a bijection that preserves all the necessary mathematical properties). The same cannot be said about infinite-dimensional spaces.

Having n distinct eigenvalues is a sufficient, but not necessary, condition for diagonalizability. The necessary condition is that, for every distinct eigenvalue, the corresponding eigenspace has the same dimension as the algebraic multiplicity of the eigenvalue. You can look up all those terms on the English wikipedia.

I'm unclear as to what help you'd like on your attached images.