# Limit of log of norm of exponential of Hamiltonian Matrix equals maximal eigenvalue

Let $$A$$ be a $$2n \times 2n$$ Hamiltonian matrix (i.e. $$JA$$ is symmetric with $$J=\begin{pmatrix} 0 & I_n \\ -I_n & 0\\ \end{pmatrix}$$). Is it true that

$$\lim_{t\to \infty}\frac{1}{t} \log \Vert{e^{At}}\Vert=\max_i(\text{Re}(\lambda_i))$$ where $$\lambda_i$$ are are all the eigenvalues of $$A$$ and $$\Vert \cdot \Vert$$ is the spectral norm?

I'm interested in computing the Lyapunov exponent of critical points of Hamiltonian systems. This type of limit appears in such cases. From numerical calculations this seems to be true.

I've been able to prove this result for the $$2\times2$$ case, where $$A=\begin{pmatrix} a & b\\ c & -a\\ \end{pmatrix}.$$ In this case, setting $$\lambda=\sqrt{a^2 + b c}$$, $$M=\frac{1}{\lambda}A$$ is an involutory matrix. Then the exponential $$\Vert{e^{\lambda M t}}\Vert$$ may be written in terms of hyperbolic trigonometric functions, which are bounded in the $$\lambda$$ imaginary case, and may be approximated in terms of exponentials in the $$\lambda$$ real case. The above result follows easily then, and it is actually independent of the norm.

I'm having problems in the general case. I wanted to find a way to write $$A$$ as a block diagonal matrix consisting of smaller Hamiltonian matrices via a unitary transformation, and then prove it inductively over $$n$$, but I've had no luck finding such result. Any suggestions?

This is true for general matrices. We can start with a Jordan block and generalize the procedure for a general Jordan decomposition.

Jordan Matrices

Consider a block $$m\times m$$ Jordan block $$B_{m,\lambda}= \lambda I_m+U_m$$, where $$I_m$$ is the $$m\times m$$ identity and $$U_m$$ is the $$m\times m$$ superdiagonal matrix on ones. Since these two matrices commute, we may write their exponential as in factored form $$\exp(tB_{m,\lambda})=\exp(\lambda t)\exp(tU_m)=e^{\lambda t}\sum_{k=0}^\infty\frac{t^k}{k!}U_m^k$$ Noting that $$U_m^m=0$$, we can rewrite $$\exp(tB_{m,\lambda})=t^{m-1}e^{\lambda t}\sum_{k=0}^{m-1}\frac{1}{k!t^{m-1-k}}U_m^k$$ This form allows us to bound the spectral norm. Choose $$t$$ such that $$\sum_{k=0}^{m-2}\frac{1}{k!t^{m-1-k}}<1$$. We can simplify the bound by noting that $$\|U_m^k\|_{sp}=1$$ for $$0\le k. $$t^{m-1}e^{\text{Re}(\lambda)t}\left(1-\sum_{k=0}^{m-2}\frac{1}{k!t^{m-1-k}}\right)<\|\exp(tB_{m,\lambda})\|_{sp} Taking the logarithm and rearranging, $$\left|\frac{1}{t}\log(\|\exp(tB_{m,\lambda}\|)-\text{Re}(\lambda)\right|<(m-1)\frac{\ln(t)}{t}-\frac{1}{t}\log\left(1-\sum_{k=0}^{m-2}\frac{1}{k!t^{m-1-k}}\right)$$ We see the right side approaches zero as $$t\to\infty$$. Thus, the expression is equal to the real part of the eigenvalue of the Jordan block.

General Matrices

Let $$\lambda_1,...,\lambda_E$$ (with be the distinct eigenvalues of $$A$$. The Jordan decomposition can be written $$A=PJP^{-1},\ \ \ J=\sum_{i=1}^E\lambda_iD_i+U$$ Where $$D_i$$ are block identity matrices and $$U$$ is a superdiagonal matrix of ones and zeros. $$U$$ does not mix the blocks of different eigenvalues, so all of these matrices commute. Exponentiating, we have $$\exp(At)=P\left(\sum_{i=1}^Ee^{\lambda_it}D_i\right)\left(\sum_{k=0}^{n-1}\frac{t^k}{k!}U^k\right)P^{-1}$$ Let $$\lambda_\max$$ be the largest real part of an eigenvalue, and $$I=\{i_1,...,i_M\}$$ be the indices such that $$\text{Re}(\lambda_{i_m})=\lambda_\max$$ be the eigenvalues of largest real Let $$p$$ be the largest power such that $$D_{i_m}U^p\neq 0$$, rearranging, we have $$\exp(At)=e^{\lambda_\max t}t^p\left(\frac{1}{p!}\sum_{m\in I}PD_{i_m}U^pP^{-1}+\sum_{i\in[1,E]\\k\in[0,n-1]\\(i,k)\notin I\times\{p\}}\frac{e^{(\lambda_i-\lambda_\max)t}t^{k-p}}{k!}PD_iU^kP^{-1}\right)$$ After isolating the $$(i_m,p)$$ term in the sum, we have factored such that the remaining nonzero terms can be made as small as necessary: if $$i\in I$$, then $$k-p<0$$ in the sum, and if $$i\notin I$$, $$\text{Re}(\lambda_i-\lambda_\max)<0$$ ensures that the term shrinks exponentially. To bound this expression, let $$\omega=\left\|\sum_{m\in I}PD_{i_m}U^pP^{-1}\right\|$$, and let $$\Omega=\max_{i,k}\|PD_iU^kP^{-1}\|$$, and choose $$t$$ such that $$\frac{\omega}{p!}>\sum_{i\in[1,E]\\k\in[0,n-1]\\(i,k)\notin I\times\{p\}}\frac{e^{\text{Re}(\lambda_i-\lambda_\max)t}t^{k-p}}{k!}\Omega$$ This gives us bounds on the norm of the exponential as before. Taking the logarithm as before, $$\left|\log\left(\frac{1}{t}\|\exp(At)\|\right)-\lambda_\max\right|<\frac{p\log(t)}{t}+\frac{1}{t}\max_{+,-}\left|\log\left(\frac{\omega}{p!}\pm\sum_{i\in[1,E]\\k\in[0,n-1]\\(i,k)\notin I\times\{p\}}\frac{e^{\text{Re}(\lambda_i-\lambda_\max)t}t^{k-p}}{k!}\Omega\right)\right|$$ By previous arguments, all of the terms in the logarithm are constant or approach zero. Therefore the limit converges.

• I've edited my answer to address the spectral norm specifically. Things are a bit more subtle there, but I think using Jordan decomposition works. – Kajelad Aug 7 at 22:31
• This time it's almost OK. It remains just to handle complex $\lambda$ (and fix the claim stated in the question ;). Two more tiny corrections are: 1) conjugation changes norms (but, yes, keeps the limit); 2) the bound after "rearranging" should respect the preceding lower bound (but, yes, the conclusion is still correct). – metamorphy Aug 8 at 6:27
• My mistake; I forgot to take the real part of $\lambda$. By "invariant under conjugation", I was referring to conjugation by an invertible matrix $A\to PDP^{-1}$. – Kajelad Aug 8 at 15:51
• @Saulpila If $A=PBP^{-1}$ then $\big|\log\lVert e^{At}\lVert-\log\lVert e^{Bt}\lVert\big|$ is bounded (by $\log(\lVert P\lVert\cdot\lVert P^{-1}\lVert)$) w.r.t. $t$. – metamorphy Aug 25 at 5:03
• @Kajelad (once again) The spectral norm of $A$ is related to the eigenvalues of $A^*A$, not $A$ itself. And those do (generally) change after a conjugation. Say, for $A=\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$, $P=\left[\begin{smallmatrix}1&0\\1&1\end{smallmatrix}\right]$ and $B=PAP^{-1}=\left[\begin{smallmatrix}-1&1\\-1&1\end{smallmatrix}\right]$, we have $\lVert A\lVert=1$ but $\lVert B\lVert=2$. – metamorphy Aug 27 at 4:59