1. ## Simple C*-algebra

Suppose we have a simple C*-algebra, say $A$. Show that $M_n(A)$ is also simple.

My first guess would be to try and obtain some contradiction by assuming $M_n(A)$ has some closed ideal $J$ other than $0$ or $M_n(A)$ and then find that $A$ will then have some closed ideal other than $0$ or $A$ which will conclude the proof.

However, I'm having some trouble with the particulars...

We can suppose that there is some sequence $\{T^{(m)}\}\in J$ that converges to $T\in J$. Then we have $RT^{(m)},T^{(m)}R\in J$ and also $RT,TR\in J$ for every $R\in M_n(A)$.

How can we now use this to construct a closed ideal in $A$? (If it is at all possible)

2. Let $e_{ij}\ (1\leqslant i,j\leqslant n)$ denote the matrix units in $M_n(A)$. Show that $e_{11}Je_{11}$ is a closed ideal in $A$ (where $A$ is identified with $e_{11}M_n(A)e_{11}$). Hence $e_{11}Je_{11} = e_{11}M_n(A)e_{11}$ (unless $e_{11}Je_{11} = \{0\}$, in which case show that $J=\{0\}$). Deduce that $e_{ij}Je_{ij} = e_{ij}M_n(A)e_{ij}$ for all i,j, and so $J = M_n(A)$.

3. I can show the first part, but I'm not sure how to attack the part where I need to show that $e_{ij}M_n(A)e_{ij}=e_{ij}Je_{ij}$?

4. Originally Posted by Mauritzvdworm
I can show the first part, but I'm not sure how to attack the part where I need to show that $e_{ij}M_n(A)e_{ij}=e_{ij}Je_{ij}$?
The (1,1)-element of the matrix $x$ is the (i,j)-element of the matrix $e_{i1}xe_{1j}$. If an element $x\in J$ can have an arbitrary element of $A$ in its (1,1)-position, then $e_{i1}xe_{1j}$ (which is also an element of $J$) can have an arbitrary element of $A$ in its (i,j)-position.