# Example that the Jordan canonical form is not “robust.”

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.

• If $\epsilon$ is small, then the change from $\epsilon$ to zero is a small change. And the effect on the Jordan form is...? – Gerry Myerson Mar 19 '13 at 5:29
• Calculate the Jordan form for $A_0$. For an arbitrarily tiny $\epsilon$ change, the Jordan form has an entry that changes from zero to one. – copper.hat Mar 19 '13 at 6:13
• Apart from the diagonal entries, can you imagine small changes in the entries of the Jordan normal form? – Marc van Leeuwen Mar 19 '13 at 7:44

## 4 Answers

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

• Time for a "nice answer" badge!! ;-) – Namaste Apr 18 '13 at 0:34

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.

I'm not sure what qualifies as large to you. Do you want the block sizes to change? Take the following matrix: $$\begin{bmatrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & \epsilon & 0\end{bmatrix}$$ If $\epsilon \neq 0$ then it's Jordan form is a single block of size $3$. If $\epsilon = 0$ then its Jordan form has a block of size $2$ and a block of size $1$.

I don't quite get the point of the example. For $2\times2$ matrices the only possible sets of sizes of blocks are $(2)$ and $(1,1)$; you can obviously go from one case to the other by (continuously) changing entries in the matrix, but I cannot see what is meant by "large changes" in the Jordan normal form. Of course when you go from $(2)$ to $(1,1)$ there is an off-diagonal entry that changes from $0$ to $1$; given that these are the only possible values, I cannot really consider that an impressive jump. And I wouldn't need much of an example to get convinced that such a jump must occur at some point.

For larger sizes of matrices, individual off-diagonal entries may still jump by $1$ at most, while the diagonal entries may be arranged to vary continuously with the matrix entries (all depends on how one orders the eigenvalues; although not easy to see it seems true that as one continuously varies the coefficients of a monic polynomial, one can choose throughout an ordering on the multi-set of its complex roots so that in each position in the list the variation is continuous (certainly not always differentiable)). But you might see more drastic jumps in the sizes of the Jordan blocks. To see this is possible, it suffices to observe that if you scale a strictly the entries of any strictly upper triangular matrix by some nonzero factor, then its Jordan normal form is unchanged, but when you reduce all entries to $0$ the Jordan type jumps to $(1,1,\ldots,1)$.