Cheng Sheng's solution is correct, but here's the typed solution
a)
Let $$P=\begin{pmatrix} -1 & -4 \\ 1 & -1\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = \det\begin{pmatrix} -1-\lambda & -4 \\ 1 & -1-\lambda\end{pmatrix}=(\lambda +1)^2 +4$$
Thus, $$\lambda_1 = -1+2i, \lambda_2 = -1-2i$$
Consider $\lambda_1 =-1+2i$.
$$N(P- (-1+2i)I) = N\begin{pmatrix} -1+1-2i & -4 \\ 1 & -1+1-2i\end{pmatrix} = N\begin{pmatrix} -2i & -4 \\ 1 & -2i\end{pmatrix}= span\{\begin{pmatrix} 2i \\ 1\end{pmatrix}\}$$
Consider $$e^{(-1+2i)t}\begin{pmatrix} 2i \\ 1\end{pmatrix} =e^{-t} (\cos(2t) + i\sin(2t))\begin{pmatrix} 2i \\ 1\end{pmatrix}\ = e^{-t}\begin{pmatrix} 2i\cos(2t) -2\sin(2t) \\ \cos(2t) + i\sin(2t) \end{pmatrix} = e^{-t}\begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} + i e^{-t}\begin{pmatrix} 2\cos(2t) \\ \sin(2t) \end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1 e^{-t}\begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} + c_2 e^{-t}\begin{pmatrix} 2\cos(2t) \\ \sin(2t) \end{pmatrix}$$
b)
As $t\to\infty$, solution asymptotically converges to the origin in counter clockwise direction in spiral shapes.
See attached image.
