I am a little stumped about something in Strangs's (great) textbook "Linear algebra and it's Applications".
At the beginning of chapter 5 (Eigenvalues, page 233) he starts off with an example that connects an eigenvalue problem to systems of differential equations.
$$ \begin{cases}\frac{dv}{dt} = 4v - 5w\\ \frac{dw}{dt} = 2v - 3w \end{cases}$$
He expresses the probelem as a matrix multiplication, clear.
$$\begin{bmatrix}dv/dt \\ dw/dt\end{bmatrix} = \begin{bmatrix}4 & -5\\2 & -3\end{bmatrix} \begin{bmatrix}v \\ w\end{bmatrix}$$
Now this is where things are confusing to me. He expresses the matrix equation using only the symbols (A, u) and then shows us how we would handle the situation if instead of matrices and vectors we had simple scalars, which would be a basic differential equation that leads to a pure exponential solution.
$$\frac{du}{dt} = \mathbf{A} \mathit{u}$$
$$\frac{du}{dt} = cu \to u(t) = e^{at}u_0$$
He then expresses the functions v and w as these pure exponential solutions, I don't understand why he can do this however because the original equations also contained terms with the other variable.
$$\begin{cases}v(t) = e^{\lambda t}y\\ u(t) = e^{\lambda t}z\end{cases}$$
He then substitutes the pure exponentials into the original system of equations and from here on out it is pretty clear.
$$\begin{cases}\lambda e^{\lambda t} y = 4e^{\lambda t}y - 5e^{\lambda t}z\\ \lambda e^{\lambda t} z = 2e^{\lambda t}y - 3e^{\lambda t}z\end{cases}$$
My question is basically, how do we get from the original equations to the pure exponentials? Why can the expressions for v(t) and w(t) be made with single terms, while the initial expressions each contain 2 terms?