TITLE: Example that the Jordan canonical form is not "robust."
QUESTION [11 upvotes]: I'm working on this problem that asks to show that the Jordan canonical form is not robust in the sense that small changes in the entries of a matrix $A$ can cause large changes in the entries of its Jordan form $J$. 
The suggestion is the consider the matrix
$$ A_\epsilon=\begin{bmatrix} \epsilon & 0 \\ 1 & 0\end{bmatrix} $$
and to see what happens to the Jordan form of $A_\epsilon$ as $\epsilon\to 0$. 
To me, the minimal polynomial of $A_\epsilon$ is then $x^2-\epsilon x$, so its eigenvalues are $0$ and $\epsilon$, and the Jordan canonical form is 
$$\begin{bmatrix} \epsilon & 0 \\ 0 & 0\end{bmatrix}. $$
But then it seems small changes in the entries of $A$ correspond to equally small changes in the Jordan form of $A$. 
Did I do something wrong? The problem is 13 of Chapter 8 in Steven Roman's Advanced Linear Algebra.

REPLY [4 votes]: When $\epsilon=0$, the Jordan form is $J_0 =\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$, when $\epsilon \neq 0$, the Jordan form is $J_\epsilon =\begin{bmatrix} \epsilon & 0 \\ 0 & 0\end{bmatrix}$. 
This may seem like a small change (as in one is 'not far' from zero), but it represents a huge structural change, and from a numerical perspective, it means that, in general, you cannot compute the Jordan form unless you are using exact arithmetic.
Generally, reasonable numerical behavior requires at least some form of continuity. The above example clearly shows that the Jordan form is not a continuous function of the input.