Originally Posted by
ymar I made a small but significant mistake in my earlier post. The vector $\displaystyle v$ is supposed to be non-zero.
The existence of an eigenvalue implies the existence of an eigenvector. With the definitions from my earlier post it's immediate. What does it mean that $\displaystyle x$ has an eigenvalue? This:
1) $\displaystyle \left (\exists \lambda\in\mathbb{C}\right)\left( \exists v\in V\setminus\{0\}\right ) xv=\lambda v.$
What does it mean that $\displaystyle x$ has an eigenvector? This:
2) $\displaystyle \left( \exists v\in V\setminus\{0\}\right )\left (\exists \lambda\in\mathbb{C}\right) xv=\lambda v.$
These formulas are equivalent. The order of the existential quantifiers doesn't matter.
However, Sheldon Axler uses a different definition of an eigenvalue. For him, that $\displaystyle x$ has an eigenvalue means this:
3) There exists $\displaystyle \lambda\in \mathbb{C}$ such that $\displaystyle x-\lambda I$ is not injective.
We want to see that 3) implies 2). What does it mean that $\displaystyle x-\lambda I$ is not injective? It means that we have two vectors $\displaystyle v_1,v_2\in V,$ such that $\displaystyle v_1\neq v_2$ and
$\displaystyle (x-\lambda I)v_1=(x-\lambda I)v_2,$
the last formula being equvalent to
$\displaystyle xv_1 -\lambda v_1 = xv_2 -\lambda v_2,$
and further equivalent to
$\displaystyle x(v_1-v_2)=\lambda (v_1-v_2).$
Since $\displaystyle v_1\neq v_2,$ we have that the vector $\displaystyle v_1-v_2$ is non-zero. Therefore, we have found a non-zero vector $\displaystyle v$ such that
$\displaystyle xv=\lambda v,$
just as we wanted.