Two days ago, my lecturer of Advanced Numerical Methods gave a review on the topic about eigenvalues and eigenvectors. Just as the lecturer presented the definition of eigenvalues and eigenvectors, a student asked the lecturer why the zero vector does not count as being an eigenvector of a square matrix. Then the lecturer replied along the line of:
The zero vector is, by definition, not an eigenvector because otherwise $\vec{A}\vec{v} = \lambda\vec{v}$ is true for any $\vec{A} \in M_{n}(\mathbb{C})$.
For some reason, I was not satisfied with that argument. In fact, I think the argument is sloppy, if not downright fallacious. However, being an engineer with no formal background in logic, I cannot disprove it. On the other hand, that reasoning was given by a professionally trained mathematician, so who am I to argue otherwise?
If I were to explain to someone else about the reason for the zero vector not being defined as an eigenvector, I would have said, based on my understanding of the problem of finding eigenvalues and eigenvectors, the following:
We would like to find a unique $\lambda$ such that the homogeneous system $(\lambda\vec{I}_{n} - \vec{A})\vec{v} = \vec{0}$ has non-trivial solution. In other words, $\lambda\vec{I}_{n} - \vec{A}$ has a non-trivial nullspace, restricting $\lambda\vec{I}_{n} - \vec{A}$ to be singular. If $\vec{v} = \vec{0}$ were allowed to be a solution, then, by the fundamental theorem of linear algebra, it had to be the only solution, i.e. $\lambda\vec{I}_{n} - \vec{A}$ had to be non-singular, contradicting our goal.
That is, the reasoning contains tangible cause-and-effect arguments. I would have also proceeded to show briefly how $\vec{v} = \vec{0}$ is a useless eigenvector when it comes to solving linear systems of ordinary differential equations, the bread and butter of science and engineering.
So the question is, is the argument given by the lecturer valid? Is it an answer commonly given by mathematics educators to such question? If it is, please enlighten me.