Algebraic Ktheory can be seen as a generalization of Linear algebra?
If yes, how so?
Algebraic Ktheory can be seen as a generalization of Linear algebra? If yes, how so? 

closed as unclear what you're asking by Stefan Kohl, quid, Alain Valette, j.c., Andy Putman Jun 8 at 23:47Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question. 


There are probably more sophisticated answers available here, but let me just mention one small way in which $K$theory relates to linear algebra, which is that the lower $K$groups successfully generalize the notion of a determinant. Let $V$ be an f.g. projective module over a ring $R$. If $V$ is of rank $n$, then its $n$th exterior power, $\Lambda^{n} V$, is a rank 1 projective $ R $ module, i.e., an element of the Picard group. Call this element $\operatorname{det}(V) $. For a short exact sequence $ 0 \rightarrow V' \rightarrow V \rightarrow V'' \rightarrow0 $ there is a canonical isomorphism $\operatorname{det}(V) \cong \operatorname{det}(V') \otimes \operatorname{det}(V'')$, so the map extends to $K_{0} (R) \rightarrow \operatorname{Pic}(R)$. The determinant picks out the nontrivial part of $ K_{0}(R)$, the part which doesn't come from free modules. Back to linear algebra. Since $\Lambda^{n} $ is a functor, if $ f:V \rightarrow V $ is a homomorphism, then $\operatorname{det} (f) := \Lambda^{n} (f)$ is in $\operatorname{End} (\Lambda^{n}V) \cong R $. In this way, the assertion $\operatorname{det}(fg) = \operatorname{det} (f) \operatorname{det}(g)$, usually a pain to prove to undergraduates, follows trivially. The map $ GL (R) \rightarrow R^{\times} $, $A \mapsto \operatorname{det}(A) $ descends to a surjection $ K_{1} (R) \rightarrow R^{\times} $, which is an isomorphism if $ R $ is a commutative local ring. The failure of this map to be an isomorphism is in some sense a measure of the failure of some parts of linear algebra over $ R $. 


The first step is to see linear algebra as encoded into vector bundles. Essentially, vector bundles are vector spaces parameterized by a topological space (or an algebraic variety). Then SerreSwan's theorem gives a correspondence between projective modules over a commutative ring and vector bundles over a compact space (or affine variety over an algebraically closed field). This gives you classical algebraic Ktheory. If you continue down this path the connection to vector spaces becomes more of a stretch; for example Waldhausen's algebraic Ktheory takes as input a category with cofibrations and weak equivalences (topologically useful classes of morphisms) and gives a spectrum as output. This seems to live only in a topological world, but classical algebraic Ktheory of a commutative ring can be realized as a special case. Though perhaps some spectra can be thought of as living in some kind of generalized linear algebraic world as well. This seems to be what Hopkins and Smith are saying in Nilpotence and Stable Homotopy Theory II
In this paper they describe the Morava Ktheory spectra $K(n)$, which have coefficients $K(n)_*\cong \mathbb{F}_p[v_n^{\pm 1}]$. These $K(n)$ act a lot like fields and this leads to some nice properties. In fact, the $K(n)$ can be thought of as the prime fields. Given that a ring spectrum $E$ is a field (i.e. $E\wedge X$ has the homotopy type of a wedge of suspensions of $X$) then $E$ has the homotopy type of a wedge of suspensions of $K(n)$ for some $n$. Perhaps the Morava Ktheories are far afield from what you had in mind, but I think there is a common thread of linear algebraic intuition behind any form of Ktheory. I think this is an interesting question (one that I have been asking as well) and I am curious what other people have to say about this. 


From here, I quote the following:
Algebraic Ktheory deals with linear algebra on a general ring; linear algebra is studied over a field. Algebraic Ktheory associates to any ring $R$ a sequence of groups $K_0R, K_1R, K_2R,\cdots$. Let me give an example of how it can be viewed as a theory of linear algebra over an arbitrary ring. Consider the $K_1$ functor. If $R$ is taken to be a field, then the calculation of $K_1R$ is simply ordinary linear algebra. (Since $K_1(R)=\mathrm{GL}(R)/[\mathrm{GL}(R),\mathrm{GL}(R)]$ (the Abelianization of $\mathrm{GL}(R)$), where $\mathrm{GL}(R)=\operatorname{colim}\mathrm{GL}(n,R)$. When $R$ is a field, it is obvious that the calculation of $K_1(R)$ reduces to ordinary linear algebra.) 

