Partition of the identity in a simple $C^*$-Algebra

Question

The following question is from $C^*$-Algebras by Examples written by Kenneth R. Davidson, Problem V.9. Given a unital simple $C^*$-Algebra $\mathfrak{A}$ and a positive element $A$, how to show there exists finitely many $\{X_i\}_{i \leq n}$ such that $\sum_{i \leq n} X_i^* A X_i = I$? Beside the original question, is it possible that all these $X_i$ are unitary?

The given hint is $X A Y + Y^* A X^* \leq X A X^* + Y^* A Y$ iff $(X - Y^*) A (X^* - Y) \geq 0$ which is true in this case. I have no idea where to come up with a $\{X_i\}_{i \leq n}$ such that $\sum_{i \leq n} X_i^* A X_i$ and how to show it is the identity. One of my attempt is to define $\mathcal{A} = \{X Z Y\,\vert\,Z \in \overline{A \mathfrak{A} A}, X, Y \in \mathfrak{A}\}$. This is an non-zero ideal and hence equal to the whole $\mathfrak{A}$.

I wish I can define $\mathcal{A}_n = \{ \sum_{i \leq n}X A X^*\,\vert\,X \in \mathfrak{A}\}$ and show that for some $n$ it would be a hereditary subalgebra. Then $0 \leq Y \leq X A X^*, \exists B \in \mathfrak{A}\,\implies\,\sqrt{Y} = B (X A X^*)^{\frac{1}{4}}\,\implies\,Y = B (X A X^*)^{\frac{1}{4}} B^* \in \mathcal{A}_n$. If this can be done, then according to the provided inequality $2 \vert\,X A Y\,\vert \leq \sum_{i \leq n} X_i^* A X_i\,\implies X A Y \in \mathcal{A}_n$ and I can replace $A$ by other elements in $\overline{A \mathfrak{A} A}$. However, it seems to me that $\mathcal{A}_n$ can hardly be an algebra... If I am on the right track where can I find such an algebra that contains finite sums of $X A X^*$?

Answer 1

It looks like you have a lot of the right ingredients, but you're looking in the wrong direction.

Let $\mathfrak I=\operatorname{span}\{XAY:X,Y\in\mathfrak A\}$. Then $\mathfrak I$ is a nonzero ideal in $\mathfrak A$, hence is dense in $\mathfrak A$ as $\mathfrak A$ is simple. But then there is some $\tilde X_k,Y_k\in\mathfrak A$ such that $\|I-\sum_{k=1}^n\tilde X_kAY_k\|<1$, so $Z=\sum_{k=1}^n \tilde X_kAY_k$ is invertible in $A$, and thus $I=Z^{-1}Z=\sum_{k=1}^n(Z^{-1}\tilde X_k)AY_k$. Now write $X_k=\frac12Z^{-1}\tilde X_k$, so that $I=2\sum_{k=1}^nX_kAY_k$. Thus $$I=\frac{I+I^*}{2}=\sum_{k=1}^nX_kAY_k+Y_k^*AX_k^*\leq\sum_{k=1}^nX_kAX_k^*+Y_k^*AY_k.$$ Now let $B=\sum_{k=1}^nX_kAX_k^*+Y_k^*AY_k$, and put $W_k=B^{-1/2}X_k$ for $k=1,\ldots,n$ and $W_k=B^{-1/2}Y_k^*$ for $k=n+1,\ldots,2n$. Then we have $$I=B^{-1/2}BB^{-1/2}=\sum_{k=1}^{2n}W_kAW_k^*.$$

As far as the second question, this fails to work for the case $\mathfrak A=\mathbb C$, as pointed out in the comments.