# Causal LTI system having exponential input

I know that for an LTI system having complex exponential input,
i.e, $$x(t)=\exp(j w_o t)$$ & $$h(t) \to$$ LTI System ;
then, its output { $$y(t) \} =M \exp(j w_o t + \phi)$$ , where $$M= |H(j w)|_{|w= w_o}$$ & $$\phi = \angle \{ H(j w)\} _{|w= w_o}$$

But for a causal LTI system having exponential input,
i.e, $$x(t)=\exp(a t)$$ & $$h(t) \to$$ Causal LTI System ;
Is it true to say its output $$\{ y(t)\}$$ is given by:
$$y(t) =K \exp(a t )$$ , where $$K= H(s)_{|s=a} \quad$$ ?

Any LTI system with transfer function $$H(s)$$ reacts to an input signal $$x(t)=e^{s_0t}$$ with an output signal $$y(t)=H(s_0)e^{s_0t}$$, if $$s_0$$ is inside the region of convergence (ROC) of $$H(s)$$. This is equivalent to saying that the functions $$e^{s_0t}$$ are eigenfunctions of LTI systems, and $$H(s_0)$$ is the corresponding eigenvalue.
Note that $$s_0$$ is an arbitrary complex constant inside the ROC of $$H(s)$$, so the relation mentioned above is also valid for $$s_0\in\mathbb{R}$$. Also note that causality is not required for the above relation to hold.
However, if your question is about input signals of the form $$x(t)=e^{at}u(t)$$, where $$u(t)$$ is the unit step function, i.e., for input signals that are zero for $$t<0$$, then the output is generally not equal to $$H(a)e^{at}u(t)$$, because the functions $$e^{at}u(t)$$ are no eigenfunctions of LTI systems.