What does it mean for a linear operator to converge to another linear operator? I am currently studying linear algebra, and I am having a trouble understanding convergence of a linear operator. Specifically,
Let $\Omega \subset L(\mathbb R^n)$ be the collection of all invertible linear operators, where $L$ stands for linear operator. Let $A, B \in \Omega$.
The book says $||A-B|| \rightarrow 0$ as $B \rightarrow A$. I am having a trouble understanding $B \rightarrow A$. Sure, $B$ and $A$ are linear operators so they can be represented by matrices form, so that I can make $||M_B - M_A||$ be very small so that perturbation will be very small. However, without resorting to thinking in terms of matrices, how can I interpret $B \rightarrow A$ in terms of linear operators?
Also, would the same conclusion hold if I am working in a linear operator in some space $X$, not necessarily Euclidean?
Thank you very much.
 A: This can be answered quite easily: Let $X, Y$ be some finite-dimensional (we are in linear algebra), normed $\mathbb{R}$-vector spaces and $A:X \rightarrow Y$ a linear operator.
Usually, one defines the so called operator norm
$$
\lVert A \rVert := \sup_{\lVert x \rVert_X = 1} \lVert Ax \rVert_Y. 
$$
It is nontrivial that $\lVert A \rVert < \infty$. This comes from Heine-Borels theorem of the unit ball being compact in finite-dimensional spaces.
This norm has some nice properties, e.g. $\lVert Ax \rVert \leq \lVert A \rVert \lVert x \rVert_X$ for all $x \in X$. There are a lot more of those and you should take a look at them.
Do also note that because $X$ and $Y$ are finite-dimensional, there exist isomorphisms $X \rightarrow \mathbb{R}^{\dim X}$ and $Y \rightarrow \mathbb{R}^{\dim Y}$. So your thinking can always be in terms of $\mathbb{R}^n$ - at least in the finite-dimensional case.
Now, some intuition: Let $\mathbb{R}^n$ have the norm $\displaystyle \lVert x \rVert_{\infty} := \max_{k = 1, ..., n} \lvert x_k \rvert$. One can show that the induced operator norm is given by
$$
\lVert A \rVert := \max_{i = 1, ..., n} \sum_{j = 1}^n \lvert  a_{ij} \rvert
$$
for some matrix $A \in \mathbb{R}^{n \times n}$. Now let's look at some matrix sequence $B^{(k)}$ that satisfies $b^{(k)}_{ij} \rightarrow b_{ij}$ for all $i, j \in \lbrace 1, ..., n \rbrace$ and some numbers $b_{ij} \in \mathbb{R}$. Let $B := (b_{ij})_{1 \leq i, j \leq n}$. Then:
$$
\lVert B^{(k)} - B \rVert = \max_{i = 1, ..., n} \sum_{j = 1}^n \left\lvert  b_{ij}^{(k)} - b_{ij} \right\rvert \overset{n \rightarrow \infty}{\longrightarrow} 0 \quad (\text{by definition})
$$
All norms on finite-dimensional spaces (so all operator norms on $\mathbb{R}^{n \times n}$) are equivalent. This means that convergence in one of them implies convergence in all the others. So a matrix converging entry-wise always implies convergence in the operator norm.
Check for yourself if the opposite is true.
If you are really interested in this then you should read a book about functional analysis. Linear operators would make a rather small portion of it.
