For a real square matrix $A$ with all eigenvalues on the imaginary axis, does the following hold?

Question

Given a real square matrix $A$ with all eigenvalues on the imaginary axis, does the following hold?

$\|e^{A}\|^2=\|Pe^{D}P^{-1}\|^2\leq\|e^{D}\|^2$ where $D$ is real Jordan form of $A=PDP^{-1}$. I know from http://www.math.wm.edu/~ckli/ima/note-3.pdf that, for unitary matrices $P$ and symmetric/anti-symmetric $D$, $\|e^{A}\|=\|e^{D}\|$. But not sure if the mentioned inequality holds when $D$ is the real Jordan normal form of $A$.

Answer 1

I found a counter-example to show that this mentioned inequality may not hold in general. Let's take $A=\begin{bmatrix}0 & 0 & 0\\0 & 0 & -2\\0 & 1 & 0\end{bmatrix}$

which has eigenvalues $0,\pm{i\sqrt{2}}$. The transformation matrix

$P=\begin{bmatrix}0 & 0 & 1\\0.8165 & 0 & 0\\0 & -0.5773 & 0\end{bmatrix}$ and the real Jordan normal form

$D=\begin{bmatrix}0 & \sqrt{2} & 0\\-\sqrt{2} & 0 & 0\\0 & 0 & 0\end{bmatrix}$.

$\|e^{A}\|=1.4084$ and $\|e^{D}\|=1$, so the inequality does not hold in general. Please correct me if I am wrong in this conclusion.

Answer 2

The eigenvalues of any matrix are invariant under conjugation. (Observe that $PMP^{-1}$ and $M$ satisfy the same minimal polynomial; it’s also easy to directly prove that an eigenvalue of one matrix is an eigenvalue of the other).

It follows that the spectral norm is invariant under conjugation, $$\|M\| = \|PMP^{-1}\|,$$ regardless of the form of $M$.