Finite subset of eigenvectors from an eigenvalue is linearly independent? Yesterday, my linear algebra professor said the following statement in class and told us to prove it as an optional homework exercise:

Statement: For a square matrix $A$ of order $n$, if $X_1,X_2,\ldots,X_r$ are any $r$ eigenvectors corresponding to an eigenvalue $\lambda$, then $X_1,X_2,\ldots,X_r$ are linearly independent.

I thought that the statement is flawed since if you consider $D$ to be an eigenvector corresponding to $\lambda$, then $\alpha D$ is also an eigenvector corresponding to $\lambda$ where $\alpha$ is any scalar.
Hence, say I choose the finite subset $\{D,2D,3D\}$ of the eigenvectors of $A$ corresponding to eigenvalue $\lambda$, then the problem statement doesn't hold.
I asked the professor about this later and he "explained" how I was misinterpreting the statement and that I don't get to choose the eigenvectors like that for myself when in fact what I did was show him a counter-example.
Since his explanation only helped himself and I'm still confused, I ask fellow MSE users to comment on whether my professor gave the statement wrong or if I'm making a mistake (if so, please explain where I went wrong).

The statement is obviously wrong. For example, if $A=I$, then for any vector $v\neq 0$, the vectors $$v, 2v, 3v,\dots, rv$$
are all eigenvectors corresponding the eigenvalue $1$, but the vectors are clearly not linearly independent.
You are correct in that. The statement clearly states "if these are eigenvectors, they are independent", and you have shown that that is not the case. Your case is logically sound.
