$\ T $ is a linear transformation. Prove $\ T + 5I $ is isomorphism

Question

$\ T: \mathbb R^3 \rightarrow \mathbb R^3 $ and $\ T(1,2,-1) = (-3,-6,3) , \ \dim(\ker T) = 2$

and I need to prove $\ T+5I $ is isomorphism.

My guess is that I should try and prove $\ \ker T \ \cap ImT = \{ 0 \} $ If I set $\ v_1 = (1,2,-1) $ to be a basis of $\ Im T $ (or can I?) and I can see $\ T(v_1) = -3 \cdot v_1 $ which mean I have one eigen value of -3 ? but then I'm stuck. in the solution it was written that $\ 0 $ is also eigen value and since $\ \dim(\ker T) = 2$ then there are two eigen vectors associated with that eigen value but I cant understand this connection.

Answer 1

Since $\dim\ker T=2$, $0$ is an eigenvalue of $T$ with multiplicity at least $2$. And you know that $-3$ is an eigenvalue of $T$ too, since $T(1,2,-1)=3(1,2,-1)$. Since $\dim\mathbb{R}^3=3$, there can't have other eigenvalues. In particular, $-5$ is not an eigenvalue. So, $T+5\operatorname{Id}_3$ is injective and therefore bijective.

Answer 2

Hint: if $\lambda$ is an eigenvalue of $T$, then $\lambda +\alpha$ is an eigenvalues of $T+\alpha I$. Then, what are the eigenvalues of your map $T+5I$? What does this tell you about its kernel?