Norm-bound for triangular Hurwitz matrix exponential

Question

Let $(-A)$ be a real Hurwitz lower-triangular matrix (this implies that all the eigenvalues of $A$ are real and negative). Since $(-A)$ is Hurwitz, we know that there exist $\alpha,\lambda>0$ such that

$$ ||e^{-At}|| \le \alpha e^{-\lambda t} $$

for all $t\ge0$. Is it possible to choose the pair $(\alpha,\lambda)$ so that $\lambda$ is equal to the smallest eigenvalue of $A$? If so, how much is the smallest corresponding $\alpha$?

Remark. From this question and the answers therein, it seems that when $(-A)$ is just Hurwitz, it is not possible to choose $\lambda$ equal to the smallest eigenvalue of $A$. But now I am introducing the further property that $A$ is lower triangular.

Answer 1

The answer to your first question is negative. The Jordan canonical form is a triangular matrix, right? That's why nothing changes.

So in general there is no best possible constant but you can take it as close as desired to the smallest eigenvalue.

Answer 2

Observe that

$$ \lambda_{\operatorname{max}}(e^{-A t}) = e^{-\gamma(A) t} \leq \lVert e^{-A t} \rVert \leq \alpha e^{-\lambda t} $$

where $\lambda_{\operatorname{max}}(X) = \max\{|\lambda| : \lambda \in \lambda(X)\}$ and $\gamma(X) := \min \{ \operatorname{Re}(\lambda): \lambda \in \lambda(X)\}$.

Since $A$ is lower triangular and real, it has real eigenvalues. Also since $-A$ is Hurwitz, it has real positive eigenvalues. Therefore, $\lambda$ in the exponent can be chosen as the minimum eigenvalue of $A$.

Also, if $A$ is diagonalizable, $\alpha$ can be chosen as $\alpha = \kappa(T) := \lVert T \rVert \lVert T^{-1} \rVert$ using the norm inequalities where $T$ is the matrix such that $T^{-1} A T$ is diagonal.

But, you can't change the exponent by selecting the constant in general. Because the exponential term is, well, exponential. So it will eventually catch the constant, independent of how big it is.

Edit: For any matrix $X \in \mathbb{C}^{n \times n}$, $\lambda_{\operatorname{max}}(X) = \rho(X) \leq \lVert X \rVert$. $\rho(X)$ is called the spectral radius of $X$. Also, if $\lambda$ is an eigenvalue of $A$, then $e^{-\lambda t}$ is an eigenvalue of $e^{-At}$ with the same eigenvector (Spectral Mapping Theorem).

Since $A$ is lower diagonal, $e^{-At}$ is also lower diagonal, because all powers of $A$ are lower diagonal. The diagonal elements of a lower diagonal matrix are also its eigenvalues, this is why $-A$ has all positive real eigenvalues.

Lastly, if $A$ is diagonalizable with $T$, then

$$ \lVert e^{-At} \rVert \leq \lVert T e^{-\Lambda t} T^{-1} \rVert \leq \lVert T \rVert \lVert e^{-\Lambda t} \rVert \lVert T^{-1} \rVert = \kappa(T) e^{-\gamma(A) t} $$

where $\Lambda$ is the diagonal matrix of eigenvalues of $A$. The last equation follows because any induced norm of a diagonal matrix is equal to its spectral radius.