Finding solutions of $Ax(t)= x'(t)$ by linear algebra Consider a system of $n$ first-order homogenous differential equations with real coefficients. We can write solutions in vector form: $Ax(t)= x'(t)$ where $A$ is the $n \times n$ coefficient matrix.  
Suppose $\lambda$ is an eigenvalue of $A$ with $\dim E_\lambda =1$ and the algebraic multiplicity of $\lambda$ equal to $2$. 
Let's guess that $y(t) =te^{\lambda t}u +e^{\lambda t}v$ is a solution to the system. I am asked to show that $u$ is an eigenvector of $A$ corresponding to $\lambda$ and that $v$ satisfies $(A-\lambda I)v=u$. Then I am asked to show that it's possible to solve for $v$.
I know that $\dim G_\lambda =2$, where $G_\lambda$ is the generalized eigenspace of $A$ corresponding to $\lambda$.
But I really have no idea how to proceed. (Even the wording of the problem confuses me.) I would appreciate some help understanding how to prove what I want.
 A: Continue with your guess $y(t) = te^{\lambda t}u + e^{\lambda t} v$, then we know for it to be a solution it must satisfy $\dot{y}(t) = Ay(t)$. Plugging our expression for $y(t)$ in gives: $\lambda e^{\lambda t}v+e^{\lambda t}u + \lambda t e^{\lambda t}u = Ate^{\lambda t}u + Ae^{\lambda t} v$. Rearrange some terms and the trick will become clear:
\begin{align}
&e^{\lambda t}(\lambda v + u) = e^{\lambda t} (Av) \iff (A-\lambda I)v=u\\
&te^{\lambda t}(Au) = te^{\lambda t}(\lambda u) \iff Au = \lambda u
\end{align}
We end up with our desired eigenvector $u$ and a generalized eigenvector $v$.
A: Suppose that $y(t) = te^{\lambda t}u + e^{\lambda t}v$ is a solution of $y^\prime(t) = Ay(t)$. Since
$$
 y^\prime(t) = e^{\lambda t}u + t\lambda e^{\lambda t}u + \lambda e^{\lambda t}v = (1+t\lambda)e^{\lambda t}u + \lambda e^{\lambda t}v,
$$
it follows that
$$
 (1+t\lambda)e^{\lambda t}u + \lambda e^{\lambda t}v = te^{\lambda t}Au + e^{\lambda t}Av
$$
for all $t$, and hence, by dividing by $e^{\lambda t}$, which is never $0$ for real $t$, and collecting powers of $t$, you get that
$$
(u+\lambda v - Av) + t(\lambda u-Au) = 0
$$
for all $t$, which, if you like, is an equality of vector-valued functions of $t$.


*

*What do you learn by setting $t = 0$ in the above equation?

*What do you learn by differentiating both sides with respect to $t$ in the above equation?

*Going back to the expression $y(t) = te^{\lambda t}u + e^{\lambda t}v$, what happens if you plug in $t = 0$? In particular, can you solve for $v$ in terms of the initial value $y(0)$?

