If you solve $(A−\lambda I) \mathbf{x}=0$ to check if $\lambda$ is an eigenvalue, and get nontrivial solutions how does it prove that $\lambda$ is indeed an eigenvalue?
In other words, why can't $\mathbf{x} = 0$ vector be a solution? And what's the reasoning between a $0$ eigenvector and invertibility?
On
If $\mathbf{x} \neq 0$, we have $Ax = \lambda x$, which is the definition of eigenvectors.
Eigenvectors are defined to be not the $0$-vector. In an eigenproblem formulation that would allow the $0$-vector to be an eigenvector, the $0$-vector would be an eigenvector for every matrix and and every value in $\mathbb{C}$ is a corresponding eigenvalue. This makes the whole concept pretty pointless.
On
By definition an eigenvalue $\lambda$ is that value such that for some $\mathbf{x}\neq 0$
$$A\mathbf{x}=\lambda \mathbf{x} \iff (A−\lambda I) \mathbf{x}=0$$
indeed we are interested to non trivial solution since wheter $\mathbf{x}=0$ we have that $A\mathbf{x}=\lambda \mathbf{x}=0$ $\forall \lambda$.
If exist $\lambda=0$ it means that $A\mathbf{x}=0$ for some $\mathbf{x}\neq 0$ then $A$ is not invertible.
Of course that the null vector is a solution. The point is: if $(a-\lambda\operatorname{Id})v=0$ for some npn-null vector $v$ (that's what non-trivial solution means here), then $v$ is an eigenvector, since $v\neq0$ and $A.v=\lambda v$.
And if $0$ is an eigenvalue, then there is a non-null vector $v$ such that $A.v=0\times v=0$. Therefore, $A$ is not invertible.