first I would like to say that I have a really bad understanding of Numerical Analysis, so the inquiry here may be annoying to some. So, if you feel you know some source you can give me, where a total newbie could begin to understand the topics in Numerical Analysis, I would be very happy.
So here is the paragraph I can't grasp:`
If E is a finite-dimensional vector space and dim(E) = n, it is known that any two norms are equivalent, and if we choose the norm $|| \ \ ||_{\infty}$, we see that the convergence of the sequence of vectors $u_k$ is equivalent to the convergence of the $n$ sequences of scalars formed by the components of these vectors (over any basis).
What is meant by any two norms being equivalent? In that textbook $u$ is defined as $Bu+c$ and $u_{k+1}=Bu_k+c$, What is c? And I am assuming that matrix $B$ is a inverse matrix of $A$? Also what is the point of multiplying B with some vector $u$ and then adding a c to get again a vector $u$? Also where do we get the first vector to compare to $\vec{u}$? I imagine that because the vector $\vec{u_k}$ must be close enough to $\vec{u}$ the components of the vectors must be close to each other as well. But in the text it is said scalars. Does it mean like scalar multiplication of each vector components? Can someone enlighten me?
`
Two norms $\parallel\cdot\parallel_1$ and $\parallel\cdot\parallel_2$ are equvialent if there are positive constants $a$ and $b$ such that for any vector $\vec{v}$, $$a\parallel\vec{v}\parallel_1\leq \parallel\vec{v}\parallel_2\leq b\parallel\vec{v}\parallel_1$$
From this it is easy to see that if a sequence of vectors converges in one of the norms, it also converges in the other, to the same value.
I don't understand the rest of your question at all. What does the sequence $u_{k+1}=Bu_k+c$ have to do with this? You say you assume $B$ is the inverse of $A$, but yuou haven't said what $A$ is.
EDIT I see that the subscripts on the norms haven't come out looking like subscripts. I'd appreciate it if someone would tell me how to format this correctly.