Finite dimensional C*-algebras Let $A$ be a finite dimensional C*-algebra. Is $A$ unital? 
I think since $A$ is finite dimensional and on finite dimensional spaces all topologies are equivalent, then $A$ is a von Neumann algebra and therefore it is unital. 
Am I right? If it's not, please give me an example of a non-unital finite dimensional C*-algebra.
 A: I think Your argument is correct - any finite dimensional $\mathrm{C}^*$-algebra is equal to it's double dual (as a Banach space), hence it has a predual - it follows that it is a von Neumann algebra. Since von Neumann algebras are unital it proves the claim. In fact more can be shown - every finite dimensional $\mathrm{C}^*$-algebra $\mathfrak{A}$ is isometrically $\ast$-isomorphism to a direct sum $\oplus_{i=1}^{n} M_{k_i}$ for some positive integers $k_{1},\dots,k_{n}$.
A: Your argument works. But you don't even need to involve topologies. Because $A$ is finite-dimensional, the spectrum of all elements consists only of eigenvalues. Then one can use functional calculus to show that all elements of $A$ are linear combinations of projections. Thinking about projections, because of finite-dimensionality every projection is a sum of minimal projections. And then, thinking first of minimal central projections and then about minimal projections in each "block", one shows that $A$ is a direct sum of full matrix blocks. In particular, it is unital. 
A: I belive an alternative argument could be:
Finite-dimensional C$^*$-algebras are fully understood: If $A$ is a C$^\ast$-algebra of finite dimension, then
$$
A\cong M_{k_1}(\mathbb{C}) \oplus M_{k_2}(\mathbb{C}) \oplus \ldots \oplus M_{k_r}(\mathbb{C})
$$
For some integer $r\geq 0$. So yes they are unital, as the right-hand side is.
