5.6 Axler's Linear Algebra Done Right Consider the following theorem:


  Theorem 5.6 Equivalent conditions to be an eigenvalue:
  Suppose V is finite-dimensional, $T \in \mathcal{L}(V),$ and $\lambda \in \mathbb{F}.$ Then the following are equivalent:
  
  (a) $\lambda$ is an eigenvalue of T ;
  
  (b) $T-\lambda I$ is not injective;
  
  (c) $T- \lambda I$ is not surjective;
  
  (d)$T- \lambda I$ is not invertible


Proof: Conditions (a) and (b) are equivalent because the equation $Tv = \lambda v$ is equivalent to the equation $(T - \lambda I)v = 0$.

From here I understand that $v \neq 0$, so that $(T - \lambda I)$ is a linear transformation that sends v to $Tv - \lambda v$. But from here I don't see why this implies that $(T - \lambda I)v = 0$ is not injective.

Can someone please explain? Thank you!
In order to tackle your question, we need the following theorems:

1- A linear map $A$ is injective, iff it has a trivial kernel. (That is, Ker($A$)={0})

2- Let $V,W$ be vector spaces, such that dim$V$ = dim$W$, $\varphi:V\rightarrow W$ a linear map. Then, the following statements are equivalent:

a. $\varphi$ is injective;

b.  $\varphi$ is surjective;

c.  $\varphi$ is bijective.

3- A map $f:X\rightarrow Y$ is bijective, iff $f$ is invertible.

Do you see how the rest of your theorem follows?
(Hint: $T-\lambda L$ is a linear transformation)
Not injective means that $\text{null}\ (T-\lambda I) \neq \{0\}$ (see 3.16). Thus $(T-\lambda I)$ is not injective if and only if there exists some $v \in \text{null}\ (T-\lambda I), v \neq 0$, so if and only if for some $v \neq 0$, $(T-\lambda I)v = 0$.