The wonders of the Spectral Theorem?

Question

Ok, I know the spectral theorem for finite dimensional linear operators states that every operator has a diagonal basis of linearly independent elements. Does it say anything deeper than this?

what i mean is that every operator is expressable as a superposition of scalars multiplied by projection matricies.

Answer 1

Yes it does.
The spectral Theorem states that, basically, there are conditions under which a matrix can be diagonalized.
However, in addition to this, or at least, as a corollary, we learn that it provides a spectral decomposition of the underlying vector space.
This is not just a statement that such a decomposition exists, rather a guide of how to perform this decomposition

Answer 2

What you are saying is false. Some operators don't have a diagonal basis. Nilpotent ones for instance. It is true however on the complex field if all the eigenvalues have multiplicity one.