The spectral norm of the truncated exponential of a matrix

Question

Let $A$ be a matrix satisfying $A^*+A\leq0$, it can be shown that $\|e^{tA}\|_2\leq1$ for all $t\geq 0$, where $\|\cdot\|_2$ is the spectral norm defined as largest singular value of the matrix.

I am curious if there is a theorem describing the spectral norm of the truncated exponential. Define $p_n(t) = I + tA + \frac{t^2}{2!}A^2 + ...+ \frac{t^n}{n!}A^n $. Is there a general theorem stating that, for what value $n$ and under what constraints for $t$ (for example, $t\|A\|_2\leq c_n$ for some constant $c_n$), as long as $A^*+A\leq0$, we still have $\|p_n(t)\|_2\leq1$?

Answer 1

The answer is already No for $n=1$. For then $$\|p_1(t)\|_2^2=\|I+tA\|_2^2=\rho((I+tA)^*(I+tA))=\rho(I+t(A^*+A)+t^2A^*A),$$ where $\rho$ denotes the spectral radius. Consider the limit case of a non-zero skew-Hermitian matrix, then $$\|p_1(t)\|_2^2=\rho(I+t^2A^*A)=1+t^2\rho(A^*A)=1+t^2\|A\|_2^2,$$ which is $>1$ for every $t>0$.

It is still No for $n=2$, because the same calculation in the same case gives $\|p_2(t)\|_2^2=1+\frac{t^4}4\|A\|_2^4$. It becomes more interesting for $n=3$ : for a skew-Hermitian matrix, one finds $$\|p_3(t)\|_2^2=\rho(I-\frac{t^4}{12}A^4-\frac{t^6}{36}A^6),$$ which is $\le1$ whenever $t\|A\|_2\le\sqrt3$. But of course, this calculation does not say anything in the general case where $A^*+A\le0$.