Elements in $K_0(A)$

Question

Let A be a $C^*$-algebra, unital or not.

1. I want to show that each element in $K_0(A)$ is of the form

$$[p]_0 - \bigg[ \begin{pmatrix} 1_n & 0_n \\ 0_n & 0_n \\ \end{pmatrix} \bigg]_0$$ for some projection $p \in M_{2n}(\tilde A)$ satisfying the following which I will call (A):

$$p - \begin{pmatrix} 1_n & 0_n \\ 0_n & 0_n \\ \end{pmatrix} \in M_{2n}(A)$$

2. And I want to show that an element $p$ in $M_{2n}(\tilde A)$ satisfies (A) if and only if $s(p)=diag(1_n , 0_n)$.

Idea:

1. By definition $K_0(A)= \lbrace [p]_0 - [s(p)]_o : p \in \mathcal{P}_\infty (\tilde A) \rbrace$ where as far as I can tell $s(a +\alpha 1)= \alpha 1$ for all $a \in A$ and all $\alpha \in \mathbb{C}$. My book says that "the image of $s_n$ is the subset $M_n(\mathbb{C}$ of $M_n(\tilde A)$ consisting of all matrices with scalar entries, and $x-s_n (x)$ belongs to $M_n(A)$ for all x in $M_n(\tilde A)$" so my question is what exactly this means. Does this mean that:

$$s(p)= \begin{pmatrix} 1_n & 0_n \\ 0_n & 0_n \\ \end{pmatrix}$$

As this seems too easy I don't think it is true. Or is there another way to show this?

2. I think that the $\Leftarrow$ should follow from the passage in my book, but I am not quite sure..

Answer 1

1. An element $x\in M_n(\tilde A)$ is a matrix of the form $x=[x_{ij}]$, where $x_{ij}\in\tilde A$ for each $i,j$, and $s_n(x)$ is the matrix $[s(x_{ij})]$. Thus we have $$x-s_n(x)=[x_{ij}-s(x_{ij})]\in M_n(A),$$ since $a-s(a)\in A$ for all $a\in\tilde A$.
   For a projection $p\in M_{n}(\tilde A)$, it isn't necessarily true that $s(p)\in M_n(\mathbb C)$ is diagonal. However, it is a projection, hence is diagonalizable. So there is some unitary $u\in M_{n}(\mathbb C)$ such that $$us(p)u^*=\begin{pmatrix}1_k&0\\0&0_{n-k}\end{pmatrix}$$ for some $k\leq n$. By adding an appropriately sized identity matrix and zero matrix, we obtain the form you're looking for.
2. Indeed, this does follow from the passage. If $s(p)=1_n\oplus0_n$, then $$p-\begin{pmatrix}1_n&0_n\\0_n&1_n\end{pmatrix}=p-s(p)\in M_{2n}(A).$$