Is there any case where there is an eigenvalue of matrix $A$ that have more than one eigenvector? (So one eigenvalue can be matched with some different eigenvectors.)
Edit: what about matrices other than zero and identity matrix?
Is there any case where there is an eigenvalue of matrix $A$ that have more than one eigenvector? (So one eigenvalue can be matched with some different eigenvectors.)
Edit: what about matrices other than zero and identity matrix?
Yes, if you take $A$ to be (for example) a $3$ dimensional identity matrix, all vectors are eigenvectors with eigenvalue $1$
For the edit, yes, it is still possible. If you (can) diagonalize the matrix and find two elements the same, you will have a repeated eigenvalue and there will be a two dimensional space of vectors with that eigenvalue. For example, $\begin {pmatrix} 2&0&0\\0&2&0\\0&0&1 \end {pmatrix}$
has all vectors in the $xy$ plane as eigenvectors with eigenvalue $2$ (plus the unit vector along $z$ with eigenvalue $1$). Now you can apply your favorite similarity transformation to make it no longer diagonal.
You should note that when talking about eigenvectors, it is more proper to rather talk about eigenspaces.
Because if $Av=\lambda v$, then $A(\alpha v)=\lambda (\alpha v)$ for any nonzero $\alpha\in\mathbb R$: every time you have an eigenvector for some eigenvalue $\lambda$, there is uncountably many of them.
But it is more interesting to think about eigenspaces. An eigenspace for some eigenvalue can have dimension greater than $1$ (as in Ross' example).
If you think of your matrix in its Jordan form, then different Jordan blocks with the same eigenvalue (as in Ross' example) are usually considered to correspond to different eigenspaces.