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If one were to read into the first chapters of a book of functional analysis, an encounter of topological concepts would be unavoidable. I found it interesting, since functional analysis can be roughly considered as the analysis of infinite dimensional vector spaces, that the same does not hold true for finite dimensional vector spaces in linear algebra.

Taking aside algebraic topology, because it mostly is not present in books for linear algebra, it seems that linear algebra is well off without the use of topological concepts.

An example: In deriving an analogue to the spectral theorem for selfadjoint matrices in finite dimensional vector spaces, one quickly discovers, that an operator simply being selfadjoint is not sufficient even for guaranteing that an eigenvalue exists. Demanding compactness of the operator then changes the situation and permits a spectrum consisting only of eigenvalues. Here we are forced to introduce a topological concept, if we want something comparable to the finite dimensional case.

Is there an enlightening way to see why the infinite cases are in need of topology and the finite cases are not? On the other hand, am I completely mistaken and just did not encounter the use of topology in linear algebra or even used it without noticing?

EpsilonDelta
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    Generally, one needs form of limit process when dealing with infinite collections of objects (cf. $\sum_k a_k$). Some notion of 'nearness' is needed. – copper.hat Jun 24 '18 at 16:53

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In elementary Linear Algebra, no topological concepts are need while starting to study finite-dimensional vector spaces, but sometimes they become useful later. Consider, for instance, this problem: how to decide whether two $n\times n$ complex matrices are similar? Of course, if they don't have the same trace or the same determinant, they cannot be similar. So, this leads to this question: is there a finite family of polynomial functions $P_k\colon M_{n\times n}(\mathbb{C})\longrightarrow\mathbb C$ ($1\leqslant k\leqslant N$) such that two matrices $A$ and $B$ are similar if and only if $P_k(A)=P_k(B)$, for each $k\in\{1,\ldots,N\}$? No, there isn't. I'll prove it when $n=2$, but the general case is similar. Since, for each $t\neq0$, $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&t\\0&0\end{smallmatrix}\right)$ are similar, we would have $P_k\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)=P_k\left(\begin{smallmatrix}0&t\\0&0\end{smallmatrix}\right)$. So,$$P_k\begin{pmatrix}0&0\\0&0\end{pmatrix}=\lim_{t\to0}P_k\begin{pmatrix}0&t\\0&0\end{pmatrix}=P_k\begin{pmatrix}0&1\\0&0\end{pmatrix},$$but $\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ are not similar.

José Carlos Santos
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Point 1) One of the powers of linear algebra is that it gives an incredibly rich theory over an arbitrary field $k$, i.e. one can fill several semesters with theory that does not need any restrictive assumption on the ground field. Now think what that means: All the basic tools and results of linear algebra hold for finite dimensional vector spaces over whatever field $k$ you like, be it $\Bbb R$, $\Bbb C$, a $p$-adic field, number fields like $\Bbb Q(\sqrt{17})$, function fields in whatever characteristic etc.

Now some of these fields come with some standard topology (actually, with much more "geometric" structure), but many just don't. And since the field $k$ basically identifies with a one-dimensional vector space over itself, already here it would seem counterproductive to come up with some topology.

As an extreme case, think of the case that the ground field is finite. Then basically the only topologies one can put on a finite dimensional vector space are the discrete and the trivial topology (which is a special case of why it's not fruitful to introduce topology on finite groups, cf. https://math.stackexchange.com/a/216566/96384). But still, many tools and results of linear algebra are applied and used in the study of finite dimensional vector spaces over these fields, which (sort of) shows that topology is just not needed for these tools and results.

On the other hand, when you start to study more advanced structure, of course the ground field and, if it has one, its topological structure become more important, even in the finite dimensional case (as in Jose Carlos Santos' answer). However, when we look at the standard fields over which that is done, we get to

Point 2) If $V$ is a finite-dimensional vector space over $\Bbb R$ and $\Bbb C$ (as well as over non-archimedean local fields), there is only one topology on $V$ which is Hausdorff and compatible with the vector space structure (i.e. addition and scalar multiplication are continuous). Cf. Is there an 'intrinsic' characterization of the usual topology on a finite-dimensional vector space? and How to endow topology on a finite dimensional topological vector space?. That topology, or actually finer geometric concepts, i.e. scalar products, norms etc., are certainly used in the more advanced linear algebra over these fields, but due to that theorem, the topological structure sometimes does not even get mentioned when in an "obvious" way one uses scaling, a unit ball etc.; however, as soon as something like this comes up, in a good linear algebra course it is pointed out that we leave the abstract theory over an arbitrary ground field $k$ and restrict to (most often) $k =\Bbb R$ and/or $k=\Bbb C$.

Torsten Schoeneberg
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