Diagonalizable endomorphism has trivial null space

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Is it correct to state that a diagonalizable endomorphism has always kernel = {0}?
 
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Is the zero operator a diagonalizable endomorphism?
 
I still don't understand.
 
R13's point was that the zero map is diagonalizable and has kernel the whole space.
 
Right, that's the case of the zero operator. But what if my eigenvalues are all non-zero (hence my eigenvectors are all linear indipendent)?
 
In that case your operator has trivial kernel.
 
Perfect. Thank you.
 
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