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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?

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    or let $A$ be the zero matrix.2012-12-09

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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.

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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.