Yes, that is trivially true. In fact any linear combination of eigenvectors for the same eigenvalue will be an eigenvector for that eigenvalue.

Another important point to appreciate is that eigenvalues and eigenvectors frequently require complex numbers. For example the eigenvalues of [[0 1] [-1 0]] are i and -i. This geometrically correlates with some kind of "rotation".

The problem arises when there is degeneration (some eigenvalues are repeated). See for example https://www.ch.ntu.edu.tw/~byjin/course/ChemMath/handouts/de...

[edit: the rest of my original comment was not correct]

If an eigenvalue is unique, the corresponding eigenvectors are on a line. If it appears twice, the corresponding eigenvectors are [edit: could be] on a plane. In general, they span [edit: could span] a subspace of dimension equal to the multiplicity.

The dimension of the space spanned by the eigenvectors associated with a degenerate eigenvalue is bounded above by the algebraic multiplicity of the eigenvalue and bounded below by one.

Not true. The matrix [[1 1] [0 1]] has the eigenvalue 1 repeated twice but the eigenvectors are on a line.

The generalized eigenvectors are on a plane.

The case where the eigenvectors do not span the space corresponds to matrices whose Jordan normal form has upper triangular bits.

You seem to have reworded to "All eigenvectors that correspond to one a unique eigenvalue lie on the same line, but have different magnitudes." This is technically true if you interpret "unique" to mean "multiplicity of one", but still seems confusing, especially as you have not yet discussed the multiplicity of eigenvalues.

I would consider rephrasing the article as you do in the above comment, "when you find an eigenvector for one eigenvalue, then you can construct infinite number of eigenvectors by scaling that one".