# B Zero eigenvalue and null space

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1. Mar 8, 2016

### 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

2. Mar 8, 2016

### Samy_A

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

3. Mar 8, 2016

### maNoFchangE

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.

4. Mar 8, 2016

### Samy_A

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.)

5. Mar 8, 2016

### maNoFchangE

Yes, it's linear.

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