Why $T+3I$ or $T-3I$ is isomorphism if $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ is linear with $\dim\ker T =2$.

Question

I have the following question :

Let $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ be a linear transformation, with $\dim\ker T = 2$ Proof that at least one of the transformation $T+3I$, $T-3I$ is isomorphism.

The only thing I could come up with is that $0$ is an eigenvalue of $T$ with geometric multiplexing of $2$, since $\dim\ker T = 2$, now I'm stuck.

Any ideas?

Thank you!

Answer 1

continue your arrgument. You have at most one more eigenvalue. Therefore at least one of the set $\{-3,3\}$ is not an eigenvalue.

Assume that $3$ is not an eigenvalue. Then for any $v\neq 0\in \mathbb{R}^3$, $Tv\neq 3v$. Hence for any $v\neq 0\in \mathbb{R}^3$, $(T-3I)v\neq 0$. Hence $T-3I$ is one to one and therefore an isomorphism.