Does $f \in \mathcal{L}(V,V)$ nilpotent and $\lambda \neq0$ imply $\lambda \operatorname{id}+f$ invertible?

Question

After proving that if $f \in \mathcal{L}(V,V)$ is nilpotent and $\lambda \neq0$ then $\lambda \operatorname{id}-f$ is invertible, I was asked if $\lambda \operatorname{id}+f$ is also invertible. My first thought was:

$$\lambda \operatorname{id}+f = -(-\lambda \operatorname{id}-f)$$

and now apply the already proven.

But the question seems to be suspiciously trivial. Does it have any infinite-dimensional trick?

It really is trivial, in particular within the realm of ring theory...but also here. — DonAntonio, Dec 10 '13 at 17:52
At first I thought it was a duplicate of http://math.stackexchange.com/questions/119904/units-and-nilpotents , but I guess it is not as it is asking about a proof strategy. — rschwieb, Dec 10 '13 at 18:04

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

Even what you wrote can be simplified.

If $f$ is nilpotent, so is $g=-f$. Now apply the first result to $g$ and $\lambda$.

You might be interested in this question if you want to see it in more general context than the ring of linear transformations.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 10 '13 at 18:02

rschwieb

153,510

Does $f \in \mathcal{L}(V,V)$ nilpotent and $\lambda \neq0$ imply $\lambda \operatorname{id}+f$ invertible?

1 Answers1