The solution of $\frac{\mathrm d\mathbf u}{\mathrm dt} = \mathbf A\mathbf u$ using eigenvectors We are given a problem
$$\frac{du}{dt} = Au$$
where 
$$ u(t) =
\left[\begin{array}{r}
y(t) \\
z(t)
\end{array}\right]
$$
and 
$$\frac{dy}{dt} = 2y - z$$
$$\frac{dz}{dt} = -y + 2z$$
hence
$$
\frac{d}{dt}
\left[\begin{array}{r}
y \\
z
\end{array}\right]=
\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]
\left[\begin{array}{r}
y \\
z
\end{array}\right]
$$
The teacher uses this example for a demonstration on the use of eigenvalues. If one represents $u(0)$ as a combination of eigenvalues of $A$, then the solution can be represented as a combination of $e^t$ and $e^{3t}$ which correspond to eigenvalues $\lambda_1 =1$ and $\lambda_2=3$. Hence
$$u(t) = Ce^t x_1+ De^{3t}x_2$$
$x_1$ and $x_2$ are the eigenvectors of $A$, and the constants $C$ and $D$ are determined by two initial values $y(0)$ and $z(0)$. 
I am curious on how to derive the result such as
$$z(t) = Ce^{t} - De^{3t}$$
I am not able to replicate this derivation, that is, how to combine eigenvectors. exponential solutions in a way that will result in $u(t)$. The eigenvalues of $A$ are 1 and 3, and eigenvectors are $$\begin{bmatrix}1\\1\end{bmatrix}\text{ and }\begin{bmatrix}1\\-1\end{bmatrix}.$$
 A: I found the answer in another lecture. This answers how seperate exponentials can combine for an overall solution. If we use $u = Sv$ where $S$ is eigenvector matrix and $v$ carries constants to combine them
$\frac{du}{dt} = Au$
$ S\frac{dv}{dt} = ASv$
$ \frac{dv}{dt} = S^{-1}ASv = \Lambda v$
using the well-known equality $S^{-1}AS = \Lambda$. Then
$ \frac{dv_1}{dt} = \lambda_1 v_1$
$ \frac{dv_2}{dt} = \lambda_2 v_2$
..
This seperates the system, each solution is
$ v_1(t) = e^{\lambda_1 t} v_1(0) $
Overall
$ v(t) = e^{\Lambda t} v(0) $
Since $u=Sv$, then $v=S^{-1}u$ and $v(0)=S^{-1}u(0)$
$S^{-1}u = e^{\Lambda t} S^{-1} u(0)$
$u = S e^{\Lambda t} S^{-1} u(0)$
The problem above takes it as far as 
$u = S e^{\Lambda t} v(0)$
If we take $v(0) = [C \ \ D]$ then
$
u(t) =
\left[\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right]
\left[\begin{array}{rr}
e^{\lambda_1 t} & \\
& e^{\lambda_2 t} 
\end{array}\right]
\left[\begin{array}{r}
C \\
D
\end{array}\right]
= 
\left[\begin{array}{r}
y(t) \\
z(t)
\end{array}\right]
$
which once calculated comes out as
$
\left[\begin{array}{r}
y(t) \\
z(t)
\end{array}\right]
=
\left[\begin{array}{r}
Ce^t + De^{3t} \\
Ce^t - De^{3t} 
\end{array}\right]
$
A: Hmm, You should have read more carefully the comments beneath your question.
Just in case you haven't read them,
you already know
$u(t) = Ce^t x_1+ De^{3t}x_2$
in vector form
$\begin{bmatrix}
       y(t)\\
       z(t)
     \end{bmatrix}
= Ce^t 
\begin{bmatrix}
       1\\
       1
     \end{bmatrix}
+ De^{3t}
\begin{bmatrix}
       1\\
       -1
     \end{bmatrix}
$
is that clear now?
you can obtain $z(t)$ by writing out the equations in coordinate-wise.

if you attempt to 'prove' why the general solution is represented in the way of the combination of eigenvectors associated with corresponding eigenvalues (for exponential).
First you should see this in the following two ways,


*

*it is a generalization of scalar differential equation(1D vector, if you'd like to call it in this way);

*it is a powerful method to solve high order DEs. it just break down a high order DE into a system of only 1st order DEs.
Now we turn back to the 1st point and considering general case(no matter what the dimension is, they look exactly same in the vector form).
Similar with scalar ODE but only written in a vector form (so please notice $u,c$ etc. are vectors)
$$\dot{u} = Au$$ 
first assume (it is true. sometimes first you guess then you prove the guess is correct. it is simply how math has been developed.)
$$u = c \cdot e^{\lambda t}$$
so for LHS
$$\dot{u} = \lambda c \cdot e^{\lambda t}$$
RHS
$$A c \cdot e^{\lambda t}$$
in order to reach equality, 
it is required 
$$\lambda c = A c$$.
That is how we end up with eigenvalues and eigenvectors. Up to now, a solution has been found. And it can be directly concluded that any eigenvector with the corresponding eigenvalue forms one solution for the DE. Then it is natrual to ask what should the general solution look like? The answer is combination of nth such terms.
Now there are also two way to explain why for n dimension system of DE we need n independent solutions.


*

*In the perspective of vector space, if the space is n dimension, a basis of the space also has to be n dimension; so in order to construct a n dim basis, you need n independent solutions.

*In the perspective of n order DE, you have nth derivatives, which implies for the explicit solution there are n constants (just think about to to find $f^{n} = 0$, you have to do n times of integration, which leads to n arbitrary constants.)
For regular case, if A have n eigenvectors, then the job is done.
Somehow, it may happen there are less than n eigenvectors due to the repeat eigenvalues, so called degenerated case. Then there is also a standard procedure to deal with this situation and you should be able to find it in any textbooks.
A: Let $\vec{u}'=A\vec{u}$ be a linear, homogeneous $n$-dimensional system of ordinary differential equations with constant coefficients. Let $\lambda, \vec{x}$ be an eigenvalue-eigenvector pair of the matrix $A$. Let
$$\vec{q}(t):=e^{\lambda t} \vec{x}.$$
Now keep in mind you can switch the order of multiplication when it comes to scalars inside matrix/vector products, and $\lambda$ and $e^{\lambda t}$ are both scalars. Also remember $\vec{x}$ is a constant vector. Then
$$ \vec{q}'(t)=\frac{de^{\lambda t}}{dt} \vec{x}$$
$$ =\lambda e^{\lambda t} \vec{x}=e^{\lambda t}(\lambda \vec{x}).$$
By definition, $A\vec{x}$ and $\lambda \vec{x}$ are equal, so
$$=e^{\lambda t}(A\vec{x})=A(e^{\lambda t}\vec{x})$$
$$ =A\vec{q}(t).$$
Hence we have $\vec{q}'(t)=A\vec{q}(t)$, so $\vec{q}$ solves the system.

Now notice that the system is linear. Thus, if $\vec{u}_1$ and $\vec{u}_2$ are two solutions to the system, and $\alpha$ and $\beta$ are any scalar constants,
$$ (\alpha\vec{u}_1+\beta\vec{u}_2)'=\alpha\vec{u}'_1+\beta\vec{u}'_2$$
$$=\alpha (A\vec{u}_1)+\beta (A\vec{u}_2) =A(\alpha\vec{u}_1+\beta\vec{u}_2).$$
Hence $\alpha\vec{u}_1+\beta\vec{u}_2$ is also a solution.

Now we put this all together. We assume $A$ is nonsingular (i.e. nonzero eigenvalues) for simplicity. Let $\lambda_1,\lambda_2,\dots,\lambda_n$ be the eigenvalues and $\vec{x}_1, \vec{x}_2,\dots,\vec{x}_n$ the eigenvectors of the matrix $A$. Furthermore, we assume that all eigenvalues are distinct (the full general solution becomes more complicated otherwise). Let $\alpha_1,\alpha_2,\dots,\alpha_n$ be arbitrary scalar constants. Then
$$\vec{u}(t)=\alpha_1 e^{\lambda_1 t}\vec{x}_1+\alpha_2 e^{\lambda_2 t}\vec{x}_2+\cdots+\alpha_n e^{\lambda_n t} \vec{x}_n$$
is a general solution to the system. In fact, given our assumptions, it is the full general solution.
