Vtyjkyuk

Let $A=fin C([0,1],M_n)mid f(0)$ is scalar matrix $$.

Then find the $K_0(A)$ and $K_1(A)$.

I am trying to use the SES $J rightarrow A rightarrow A/J$ where $J$ can be taken as some closed ideals of $A$. Such a closed ideals may be found in a similar way due to my previously asked question here. Now $J$ contains all the continuous functions that are zero on some closed subsets of $[0,1]$, that is, finite points or the closed subintervals of $[0,1]$. Then it seems to me that $M_n(J)$ consists of those matrices with operator-entries that is actually zero in above case. These matrices have zero diagonals and therefore $K_0(J)=0$. But after that I don't know how to do apart from $K_0(A)simeq K_(A/J)$ by the half-exactness of SES.

I am also assuming that there should be a way to consider the $K$-groups by definition, for example, when calculating $K(mathbb C)$, take the difference of $[p]_0-[q]_0$, each of which is the equivalence class of matrices in $M_n(mathbb C)$. Then the generators are just those matrices with different numbers of $1$ in the diagonal, plus all zeros one; so $V(A)$= $mathbbN cup 0$. Counting the difference, we have $K_0(mathbbC)=mathbb Z$.

Could anybody show me the technique for calculating $K_1$ and $K_0$? Are there any standard techniques to calculate $K$-groups in general? Thanks.

$requireAMScd$
If you want to use six-term exact sequence, one can consider the followings short exact sequence:
$$
0 to CM_n overset imath to A overset mathrm ev to M_n to 0,
$$
where $CM_n$ is the cone over $M_n$, $imath$ the inclusion and $mathrmev$ the evaluation map at $0$.

beginCD
K_0(CM_n) @>> > K_0(A) @>> > K_0(M_n)\
@AAA @. @VVV \
K_1(M_n) @<< < K_1(A) @<< < K_1(CM_n)
endCD
Since the cone over any algebra is contractible, and $K_1(M_n)$ is trivial, this becomes

beginCD
0 @>> > K_0(A) @>> > K_0(M_n)\
@AAA @. @VVV \
0 @<< < K_1(A) @<< < 0
endCD

So we have an isomorphism
$$
K_0(mathrmev) colon K_0(A) to K_0(M_n) cong mathbb Z
$$
and $K_1(A) = 0$.

Remark:
The same proof works if you look at functions $f colon [0,1] to M_n$ such that $f(0)$ is diagonal.

One can describe the algebra $A$ as the unitization of $M_n(B)$ where $B = C_0[0,1)$. Hence,
$$
K_0(A) cong K_0(M_n(B))oplus K_0(mathbbC)
$$
But $K_0(M_n(B)) cong K_0(B)$ by the stability of $K_0$, and $K_0(B) = 0$ since the identity map on $B$ is homotopic to the zero map. Thus,
$$
K_0(A)cong mathbbZ
$$