The following question was part of a problem sheet in numerical maths:
Given the scalar equation $\ddot{y} = - \lambda y$, \lambda > 0
show that for h > h_{critical}, the approximate solution goes to infinity as $t$ goes to infinity. This behaviour is called blow-up.
The Stormer-Verlet-method is $y_{k+1} = -y_{k-1} + 2y_k - h^2 \lambda y_k$, which can be rewritten as
$ \left( \begin{array}{c} y_{k+1} \\ y_k \end{array} \right) = \left( \begin{array}{cc} 2 - h^2 \lambda & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} y_{k} \\ y_{k-1} \end{array} \right)$
Then in the solution they write that there is no blow-up if and only if the eigenvalues $\mu_i$ of the matrix $ B = \left( \begin{array}{cc} 2 - h^2 \lambda & -1 \\ 1 & 0 \end{array} \right)$ are all $\leq 1$.
My question is: how do they get to this claim about the relationship between eigenvalues and blow-up?
I read that the 2 norm $|| . ||_2$ is the same as the spectral norm if the matrices are square. Then I was wondering if the reason could be something like $ || Av||_2 \leq || A ||_2 || v ||_2$ and then using that this is the spectral norm: $ || Av||_2 \leq || A ||_2 || v ||_2 \leq ( max_i \mu_i ) || v ||_2$ where the max is taken over the eigenvalues of $A$.
The problem with this is, that I need the other direction: $ ( max_i \mu_i ) || v ||_2 \leq || Av||_2 $
Then, obviously, if the biggest eigenvalue is bigger than 1, the approximation after $k$ steps is $A^k y_0$ and using the above I get a blow-up.
Many thanks for your help.