Is the Null Space of an Operator Defined by Its Zero Eigenvalue?

[maNoFchangE]

Suppose $T$ is an operator in a finite dimensional complex vector space and it has a zero eigenvalue. If $v$ is the corresponding eigenvector, then
$$
Tv=0v=0
$$
Does it mean then that $\textrm{null }T$ consists of all eigenvectors with the zero eigenvalue?
What if $T$ does not have zero eigenvalue? Does it mean that its null space is just the zero vector?

Thanks

 

[Samy_A]

Suppose $T$ is an operator in a finite dimensional complex vector space and it has a zero eigenvalue. If $v$ is the corresponding eigenvector, then
$$
Tv=0v=0
$$
Does it mean then that $\textrm{null }T$ consists of all eigenvectors with the zero eigenvalue?
What if $T$ does not have zero eigenvalue? Does it mean that its null space is just the zero vector?

Thanks

Yes.
(Except that $\textrm{null }T$ consists of all eigenvectors with the zero eigenvalue and the 0 vector.)

 

[maNoFchangE]

Yes.

So, the answer to all of my questions is affirmative?
Then, if $T$ does not have a zero eigenvalue, it's equivalent of being injective.

 

[Samy_A]

So, the answer to all of my questions is affirmative?
Then, if $T$ does not have a zero eigenvalue, it's equivalent of being injective.

For a linear operator, yes.
(I should have mentioned that in my first answer too, I just assumed you meant that T is a linear operator.)

 

[maNoFchangE]

Yes, it's linear.