Relation between Eigenspaces of Similar Matrices This is a follow up question to, Eigenvectors of a matrix and its diagonalization.
Do the eigenspaces corresponding to the same eigenvalues of similar matrices describe the same subspaces?  For example, let's say I have two similar matrices $A,D\in \mathbb{R}^3$ and that $D$ is diagonal.  Suppose $\lambda_1$ is a unique eigenvalue of both and that  $V_{\lambda_1}$ is an eigenspace of $A$ describing a line in $\mathbb{R}^3$.  Does the eigenspace of $D$ corresponding to $\lambda_1$ describe the same line?
 A: You have to distinguish between the "underlying" linear transformation and the representation.
Recall that similar matrices $A$ and $B$ may be thought of as representing the same linear transformation with respect to different basis. What this means, explicitly, is that we have an automorphism (a bijective map that respects sums and scalar multiplications) $P\colon\mathbb{R}^n\to\mathbb{R}^n$, with the property that for every vector $\mathbf{v}$, $P(A\mathbf{v})) = B(P(\mathbf{v}))$. 
You can think of $P$ as a "dictionary" that from $\beta$-speak to $\gamma$-speak, where $\beta$ and $\gamma$ are two distinct basis, in the sense that if $[\mathbf{v}]_{\beta}$ is the coordinate vector of $\mathbf{v}$ with respect to $\beta$, and $[\mathbf{v}]_{\gamma}$ is the coordinate vector of $\mathbf{v}$ with respect to $\gamma$, then $P[\mathbf{v}]_{\beta}=[\mathbf{v}]_{\gamma}$.
In that sense, if you look at the underlying vectors (not at the coordinate vectors), then the eigenvectors of similar matrices are the same vector: after all, $A$ and $B$ represent the same linear transformation, it's just that $A$ "speaks" $\beta$-language and $B$ "speaks" $\gamma$-language. It's just like saying that The Odyssey in English is the same as The Odyssey in Russian: it's the same story, with the same characters, doing the same things. 
On the other hand, if you look at the coordinate vectors, so that you view each of $A$ and $B$ as simply operating on $\mathbb{R}^n$ with the standard basis, then the eigenspaces need not be the same; for instance, the matrices
$$A=\left(\begin{array}{cc}1&1\\1&1
\end{array}\right)\quad\text{and}\quad B=\left(\begin{array}{cc}2&0\\0&0\end{array}\right)$$
are similar, via $P^{-1}AP = B$ with
$$P=\left(\begin{array}{rr}1&1\\
1 & -1\end{array}\right),$$
but the eigenspaces of $A$ are $\{(a,a)\mid a\in\mathbb{R}\}$ and $\{(b,-b)\mid b\in\mathbb{R}\}$, while the eigenspaces of $B$ are $\{(a,0)\mid a\in\mathbb{R}\}$ and $\{(0,b)\mid b\in\mathbb{R}\}$. Following the analogy from above, it's because on their face, The Odyssey in English is not the same as The Odyssey in Russian: different alphabets, different words, different sentence structure, etc. 
So it depends how you want to view the matrices. As operators on $\mathbb{R}^n$, no, they don't have to have the same eigenspaces. As coordinate matrices of a particular linear transformation with respect to different bases, then yes, they have the same eigenspaces... but they describe them differently: one uses coordinate vectors with respect to one basis, the other matrix uses coordinate vectors with respect to a different matrix.
