General solution to $u'=\left(\begin{smallmatrix} -1 & 1\\ 0 & -1 \end{smallmatrix}\right)u$ How to find the general solution to the following linear system:
$$\left(\begin{matrix}
u_1'\\
u_2'
\end{matrix}\right)=\left(\begin{matrix}
-1 & 1\\
0 & -1
\end{matrix}\right)
\left(\begin{matrix}
u_1\\
u_2
\end{matrix}\right)$$
My working so far: The characteristic polynomial of $A$ is given by
$$p(\lambda)=(-1-\lambda)^2$$
giving an eigenvalue of $\lambda =-1$ of algebraic multiplicity 2. To find the eigenvectors corresponding to this eigenvalue we compute the nullspace of $A-\lambda I$, so
$$N(A+I)=N\left(\begin{matrix}
0&1\\
0&0
\end{matrix}\right)=
\left(\begin{matrix}
1\\
0
\end{matrix}\right)=\overline{v}_1$$
So we find that the eigenvalue $\lambda=-1$ has algebraic multiplicity 2 and geometric multiplicity of only 1. Now because there is only one eigenvector I computed the generalized eigenvector via 
$$N((A+I)^2)=N\left(\begin{matrix}
0&0\\
0&0
\end{matrix}\right)=\left(\begin{matrix}
1\\
1 \end{matrix} \right)= \overline{v}_2$$
Now I think the general solution is not
$$y=c_1\overline{v}_1e^{-x}+c_2\overline{v}_2e^{-x}$$
But I am unsure. So this is how far I got. How do I finish my arguments to find the general solution, did I even need to find the generalized eigenvector? Thanks
 A: You only have to compute 
$$
e^{tA}=\sum_{n=0}^\infty\frac{t^n}{n!}A^n, 
$$
where
$$
A=\left(\begin{array}{cc}-1&1\\0&-1\end{array}\right).
$$
We have
\begin{eqnarray}
A^2&=&\left(\begin{array}{cc}-1&1\\0&-1\end{array}\right)\cdot\left(\begin{array}{cc}-1&1\\0&-1\end{array}\right)=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right),\\
A^3&=&\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)\cdot\left(\begin{array}{cc}-1&1\\0&-1\end{array}\right)=\left(\begin{array}{cc}-1&3\\0&-1\end{array}\right).
\end{eqnarray}
If for all $k=1,2,\ldots, n$ we assume that 
$$
A^k=\left(\begin{array}{cc}(-1)^k&(-1)^{k-1}k\\0&(-1)^k\end{array}\right),
$$
then
$$
A^{n+1}=A^nA=\left(\begin{array}{cc}(-1)^n&(-1)^{n-1}n\\0&(-1)^n\end{array}\right)\cdot\left(\begin{array}{cc}-1&1\\0&1\end{array}\right)=\left(\begin{array}{cc}(-1)^{n+1}&(-1)^n(n+1)\\0&(-1)^{n+1}\end{array}\right).
$$
Hence
$$
A^n=\left(\begin{array}{cc}(-1)^n&(-1)^{n-1}n\\0&(-1)^n\end{array}\right)\quad \forall n \in \mathbb{N}.
$$
It follows that
$$
e^{tA}=\sum_{n=0}^\infty\frac{t^n}{n!}\left(\begin{array}{cc}(-1)^n&(-1)^{n-1}n\\0&(-1)^n\end{array}\right)=\left(\begin{array}{cc}\sum_{n=0}^\infty\frac{(-t)^n}{n!}&t\sum_{n=1}^\infty\frac{(-t)^{n-1}}{(n-1)!}\\0&\sum_{n=0}^\infty\frac{(-t)^n}{n!}\end{array}\right)=\left(\begin{array}{cc}e^{-t}&te^{-t}\\0&e^{-t}\end{array}\right).
$$
Hence
$$
u(t)=\left(\begin{array}{cc}e^{-t}&te^{-t}\\0&e^{-t}\end{array}\right)u_0=e^{-t}\left(\begin{array}{cc}1&t\\0&1\end{array}\right)u_0=e^{-t}{a+bt\choose b} \text{ with } u_0={a\choose b} \in \mathbb{R}^2.
$$
Edit:
An easier way to compute $e^{tA}$ is to use the fact that the matrix $B:=A+I=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$ satisfies $B^n=0$ for every $n\ge 2$ (which follows from the Cayley-Hamilton Theorem). Thus
$$
e^{tA}=e^{tB-tI}=e^{-tI}e^{tB}=e^{-t}(I+tB)=\left(\begin{array}{cc}e^{-t}&te^{-t}\\0&e^{-t}\end{array}\right).
$$
