Why these are equivalent?

Question

Situation: operator theory, spectrum of a operator.

We consider this as definition:

$\lambda$ is a eigenvalue if $\lambda x=Tx$ for some $x\ne 0$

but I see someone saying this:

$\lambda x-Tx=0\not \Rightarrow x=0 $ so $\lambda $ is a eigenvalue.

I cannot see why the latest sentence implies $\lambda$ is a eigenvalue. Some help? Is this a very basic logic problem?

Answer 1

The statement "$\lambda x = Tx$ for some nonzero $x$" is the same as "$\lambda x - Tx = 0$ for some nonzero $x$." So if $\lambda x - Tx$ doesn't imply $x = 0$, then there's a nonzero $x$ satisfying the equation, so you're back to the first statement.

It might help to see that both formulations are equivalent to the third formulation

$\lambda$ is an eigenvalue iff $\lambda I - T$ has a nontrivial kernel.

Answer 2

Note that $P\not\Rightarrow Q$ is the same as $P\land\neg Q$. But still this alternative formulation is ugly in my opinion as one must be careful with the implicit quantifiers.