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.
Best Answer
Lets see if we can do a few examples so you can get a better feel.
Lets write the Jordan Normal Form for three different cases.
Example 1: General Case
$A_{\epsilon} = \begin{bmatrix} \epsilon & 0 \\ 1 & 0\end{bmatrix} = P J P^{-1} = \begin{bmatrix} 0 & \epsilon\\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix} 0 & 0 \\ 0 & \epsilon \\ \end{bmatrix} \cdot \begin{bmatrix} -\frac{1}{\epsilon} & 1 \\ \frac{1}{\epsilon} & 0 \end{bmatrix}$
Example 2: $\epsilon = 1/10000$
$A_{\epsilon} = \begin{bmatrix} \frac{1}{10000} & 0 \\ 1 & 0\end{bmatrix} = P J P^{-1} = \begin{bmatrix} 0 & \frac{1}{10000}\\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix} 0 & 0 \\ 0 & \frac{1}{10000} \\ \end{bmatrix} \cdot \begin{bmatrix} -10000 & 1 \\ 10000 & 0 \end{bmatrix}$
Example 3: $\epsilon = 0$
$A_{\epsilon} = \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} = P J P^{-1} = \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Compare the Jordan block, $J$ in these three cases and what do you notice?
Now, generalize the argument using limits and matrices.
Regards