differential equations in linear algebra I can't seem to grok how this differential is working, I think I have missed something simple along the way but IDK where. In the book "Linear Algebra and Its Applications" by Gilbert Strang at the beginning of chapter 5 he is briefly going over differential equations and the following formulas are given...
$$
\frac{dv}{dt} = 4v-5w \quad v=8 \quad at \quad t=0 \\
\frac{dw}{dt} = 2v - 3w \quad w=5 \quad at \quad t = 0 
$$
With the vector unknowns
$$
u(t) = \begin{bmatrix}v(t) \\ w(t) \end{bmatrix} \quad
u(0) = \begin{bmatrix} 8 \\ 5 \end{bmatrix} \quad
A = \begin{bmatrix}4 & -5 \\ 2 & -3 \end{bmatrix}
$$
One line later it says
$$
\frac{du}{dt} = Au \quad with \quad u = u(0) \quad at \quad t = 0
$$
I follow the first part as 


*

*derivative of function $v \;\; w.r.t \;\; t$, as well as derivative of function $w \;\; w.r.t. \;\; t$

*now we have a vector of unknown functions that we know the derivative of, and the starting point. 

*The derivative of our unknown functions should be... $A$, right??? How do we get $Au$ as the derivative?

 A: $A$ is the matrix of the constant coefficients of the matrix differential equation. It it just a matter of notation. 
The derivative of $u$ in terms of $u$ is described by $A$ if you will. Or put in other words, the derivative of each component is a sum of the components and the weights are contained in $A$.
A: Your misunderstanding is in (3). The derivative of the vector function $u$ is another vector function: in coordinate terms, 
$$
u' = (v', w').
$$
The derivative can't be the matrix $A$. That's just an array of numbers. The equation
$$
u' = Au
$$
 tells you how to combine the derivatives of $v$ and $w$ to get the derivative of $u$.
A: It is just a simple way of writing nothing more, if you multiply $A$ by $u$ you'll get $Au=[4v-5w,2v-3w]^t$ and from what is above you know that $\frac{dv}{dt}=4v-5w$ and  $\frac{dw}{dt}=2v-3w$
So by substitution you find out that $Au=\begin{bmatrix} \frac{dv}{dt} \\ \frac{dw}{dt} \end{bmatrix}=\begin{bmatrix} v'(t) \\ w'(t) \end{bmatrix}$
Also, at $t=0$ $$u=u(0)=\begin{bmatrix} v(0) \\ w(0) \end{bmatrix}=\begin{bmatrix} 8 \\ 5 \end{bmatrix}$$
A: The system is indeed
$$\dot u=\begin{pmatrix}\dfrac{dv}{dt}\\\dfrac{dw}{dt}\end{pmatrix}=\begin{pmatrix}4&-5\\2&-3\end{pmatrix}\begin{pmatrix}v\\w\end{pmatrix}=Au.$$
