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!
Best Answer
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)