Can the condition number of a Jordan basis be made stubbornly large?

Question

For each $k \in \mathbb R$, does there exist a non-empty open ball $B$ of $\mathbb R^{2 \times 2}$ such that for all $M \in B$ and Jordan decompositions $PJP^{-1}$ of $M$, the condition number $\kappa(P)$ is greater than $k$?

The condition number of $P$, denoted $\kappa(P)$, is equal to $\lVert P \rVert \lVert P^{-1} \rVert$, where we're using the operator norm.

If it's true then it suggests that the problem of having an ill-conditioned Jordan basis of $M$ cannot be solved by perturbing $M$ by an arbitrarily small distance $\varepsilon$. The perturbation might need to be by a minimum distance. We can say that it's stubbornly large.

Note that the term "Jordan decomposition" is ambiguous. I refer to the Jordan Normal Form in linear algebra. The problem is, I think the term Jordan Normal Form could be misunderstood to mean just the $J$ factor, and not the $P$ factor which is needed to describe such a decomposition. The term Jordan decomposition sometimes refers to a different decomposition which is not the JNF.

Answer 1

Yes. Let $\kappa > 1$ and consider the matrix $$ A = \begin{bmatrix} -\kappa^2+4\kappa-1 & \kappa^2-1 \\ -\kappa^2 + 1 & \kappa^2+4\kappa+1 \end{bmatrix} = P D P^{-1} $$ where $$ P = \alpha \begin{bmatrix} \kappa+1 & \beta (\kappa - 1) \\ \kappa-1 & \beta (\kappa+1) \end{bmatrix}, \quad D =\begin{bmatrix} 2 \kappa & 0 \\ 0 & 6 \kappa \end{bmatrix} $$ for any nonzero scalars $\alpha, \beta$. Note that the $P$ factor is unique in this case only up to an arbitrary column scaling (here represented by $\alpha$ and $\beta$), and the condition number of $P$ is actually ambiguous. Nonetheless, in this case the condition number (in the 2-norm) of $P$ is at least $\kappa$, attained when $|\beta| = 1$. The conclusion now follows by taking, say, $\kappa = 2 \max(k,1)$, and noting that the eigenvectors are continuous at $A$: for fixed $B$, $A + \epsilon B$ has eigenvectors $$ \begin{pmatrix} \kappa+1+O(\epsilon) \\ \kappa-1 \end{pmatrix}\text{ and }\begin{pmatrix} \kappa-1+O(\epsilon) \\ \kappa+1 \end{pmatrix}$$ as $\epsilon \to 0$.

Edit: Using the choice of eigenvectors above together with an arbitrary column scaling results in a matrix $P$ that is continuous as a function of the matrix $A$ (at $A$), and continuous in the choice of column scalings, and it follows that the minimum condition number of $P$ over all choices of column scalings (covering all possible $P$) is also continuous at $A$.

At $A$ (i.e., for $\epsilon = 0$), we can see that the minimum condition number of $P$ is $\kappa$ as follows. The squares of the singular values of $P$ are the eigenvalues of $P^*P$, given by $$ \sigma^2 = |\alpha|^2 (\kappa^2 + 1) (a \pm \sqrt b) $$ where $$ a = 1+|\beta|^2, \quad b = \left(|\beta|^2 + \frac{\kappa^4-6\kappa^2+1}{(\kappa^2+1)^2}\right)^2 + 16 \kappa^2 \frac{(\kappa^2-1)^2}{(\kappa^2+1)^4} . $$ Note $b > 0$ and $a^2 - b = 16 |\beta|^2 \kappa^2 / (\kappa^2+1)^2 \ge 0$. The square of the condition number of $P$ is the ratio of squares of the singular values, $$ \kappa(P)^2 = \frac{a+\sqrt{b}}{a-\sqrt{b}} . $$ Letting $x = |\beta|^2$, the logarithmic derivative of $\kappa(P)^2$ is $$ \frac{1}{\kappa(P)^2} \frac{d}{dx} \kappa(P)^2 = \frac{1}{\sqrt{b}} \frac{a \frac{db}{dx} - 2b \frac{da}{dx}}{a^2-b} = \frac{1}{\sqrt{b}} \frac{x-1}{x} . $$ So, $\kappa(P)$ is decreasing as a function of $x = |\beta|^2$ for $x < 1$ and increasing for $x > 1$, and the minimum occurs at $x = 1$, i.e., $|\beta| = 1$, where $\kappa(P) = \kappa$.