Consider the real square matrix $A \in\mathbb{R}^{2n,2n}$ of the form:

$$ A= \begin{bmatrix} 0_n & I_n \\ -P & 0_n \end{bmatrix}$$ where $P$ is a tridiagonal symmetric positive-definite matrix.

Then define $B=A-E_{2n,n-1}$ where $E_{2n,n-1}$ has all elements zero except $1$ at $(2n,n-1)$.

I am trying to express $e^{tB}$ as a function of $e^{tA}$ for $t\in\mathbb{R}$. Unfortunately, $A$ and $E_{2n,n-1}$ do not commute so the exponential of the sum is not the product of the exponential. $[A,E_{2n,n-1}]$ does not commute with $A$ so the Zassenhaus formula is not very helpful (cf Not commuting exponential matrices).

However, given the simple expressions of the matrices (especially $E_{2n,n-1}$), I still have hope that $e^{Bt}$ can be expressed in terms of $e^{At}$. If it helps, $P$ can be chosen as:

$$P=\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix}$$

So the question if there's an explicit relationship between $e^{At}$ and $e^{Bt}$. If you think there's none, I am also interested in understanding why.


1 Answer 1


Do you mean $n=3$ and $$ A = \left[ \begin {array}{cccccc} 0&0&0&1&0&0\\ 0&0&0&0 &1&0\\ 0&0&0&0&0&1\\ -2&1&0&0&0&0 \\ 1&-2&1&0&0&0\\ 0&1&-1&0&0&0 \end {array} \right],\ B = \left[ \begin {array}{cccccc} 0&0&0&1&0&0\\ 0&0&0&0 &1&0\\ 0&0&0&0&0&1\\ -2&1&0&0&0&0 \\ 1&-2&1&0&0&0\\ 0&0&-1&0&0&0 \end {array} \right] ? $$ Since $B$'s characteristic polynomial factors as $(\lambda^2+3)(\lambda^2+1)^2$, its matrix exponential is rather simple in form, though moderately complicated to write out: each entry is a linear combination of $\cos(t)$, $t \cos(t)$, $\sin(t)$, $t \sin(t)$, $\cos(\sqrt{3}t)$ and $\sin(\sqrt{3}t)$.

On the other hand, $A$'s characteristic polynomial is $\lambda^6 + 5 \lambda^4 + 6 \lambda^2 + 1$ which is irreducible over the rationals. All eigenvalues are simple (and imaginary). The matrix elements of $\exp(At)$ are linear combinations of $\cos(\omega_i t)$ and $\sin(\omega_i t)$ where $-\omega_i^2$ are the roots of $x^3 + 5 x^2 + 6 x + 1$.

Now $\exp(Bt)$ can't be a continuous function of $\exp(At)$. $\exp(At)$ is bounded, and for any $\delta > 0$ there are arbitrarily large $t$ for which $\|\exp(At) - I\| < \delta$. On the other hand, because of the $t \cos(t)$ and $t \sin(t)$ terms, $\|\exp(Bt)\| \to \infty$ as $t \to \infty$.

  • $\begingroup$ Yes, $A$ and $B$ are correct. Thank you for your answer. I understand that $\exp(At)$ is bounded and not $\exp(Bt)$, but why couldn't $\exp(Bt)$ be a continuous function of $\exp(At)$? For example, $x\mapsto 2x$ is a continuous function of $x\mapsto 2$, but one is bounded and not the other. $\endgroup$
    – anderstood
    Sep 30, 2016 at 23:49
  • $\begingroup$ Suppose there was a function $f$, continuous at $I = \exp(A0)$, such that $\exp(Bt) = f(\exp(At))$. Continuity says that for any $\epsilon > 0$, there is $\delta > 0$ such that $\|\exp(At) - I\| < \delta$ implies $\|\exp(Bt) - I\| < \epsilon$. But as I said, there are arbitrarily large $t$ for which $\|\exp(At)-I\| < \delta$, while $\|\exp(Bt)-I\| < \epsilon$ is only true for a bounded set of $t$. $\endgroup$ Oct 1, 2016 at 0:02
  • $\begingroup$ I was mixing the continuity of $f$ w.r.t $t$ and $w.r.t$ it's matrix argument. That's the second one you're talking about, if I'm correct. $\endgroup$
    – anderstood
    Oct 1, 2016 at 0:18

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