If the eigenvalues of a complex matrix have negative real parts, then the zero solution of the corresponding linear system of first order differential equations is globally asymptotically stable

Consider the first order differential equation (1) $\frac{d}{dt} Y = AY$, where $A$ is an $n \times n$ complex matrix. A solution of (1) (indeed, of any differential equation) is called a steady state if it is constant in $t$. A steady state $C$ is called globally asymptotically stable if for any other solution $Y$, we have $\mathsf{lim}_{t \rightarrow \infty} Y(t) = C$. (Limits are taken entrywise.) See pages 507-508 of D&F for a more lucid explanation of steady states and globally asymptotically stable steady states.

Prove that if the eigenvalues of $A$ have negative real parts, then the zero solution of $\frac{d}{dt} Y = AY$ is globally asymptotically stable.

[My usual disclaimer about analytical problems applies. Be on the lookout for Blatantly Silly Statements.]

As we have seen, every solution of (1) is a linear combination of the columns of $\mathsf{exp}(At)$. If $P^{-1}AP = B$ is in Jordan canonical form, then every solution of (1) is a linear combination of the columns of $P\mathsf{exp}(Jt)$. Note the exponential of a Jordan block as we computed here; in particular, we can see that if $Y$ is a solution of (1), then each entry of $Y$ is a linear combination of functions of the form $p(t)e^{\lambda t}$, where $\lambda$ is an eigenvalue of $A$ and $p(t)$ is a polynomial in $t$. Say $\lambda = a+bi$; then $e^{\lambda t} = e^{at}(\mathsf{cos}(bt) + i \mathsf{sin}(bt))$.

[Start handwaving]

Since $a$ is negative, as $t$ approaches infinity, $e^{at}$ tends to 0 faster than any polynomial, so that our solution $Y$ tends to the constant solution 0.

[End handwaving]

Thus the zero solution is globally asymptotically stable.