Question:Let $A$ be a $C^*$-algebra, $a \in A$ self adjoint. Suppose that the spectrum $\sigma(a)$ is an infinite set. Show that $A$ is infinite-dimensional.
How can i prove it?
I guess: Let $A$ be finite dimensional. Thus it is isomorphic with direct sum of matrixes $(n_i)$.$\sigma(a)$ is eigenvalue of this direct sum. Can I say then that $\sigma(a)$ is finite? Is it true?
On
Assume $k$ field and $A$ a finite dimensional $k$ algebra. Then the spectrum of any element is finite.
Step 1. Every element $a$ of $A$ satisfies a polynomial equation of degree at most $n= \dim_{k} A$. Indeed, the elements $1$, $a$, $\ldots$, $a^n$ are linearly dependent.
Step 2. Let $a$ satisfying a polynomial equation $P(a) = 0$, where $P \in k[X]$ , $P\ne 0$. Then the spectrum of $a$ is included in the zero set of $P$, so has cardinality at most $\deg P$, hence finite.
Indeed, let $\sigma \in k$ so that $P(\sigma) \ne 0$. We have the division with remainder $$P(X) = (X-\sigma) \cdot Q(X) + P(\sigma)$$
Plug $a$ instead of $X$ above and get $$0 = P(a) = (a - \sigma ) \cdot Q(a) + P(\sigma)$$ so $(a-\sigma)^{-1}$ is invertible with inverse $- P(\sigma)^{-1} \cdot Q(a)$
From the above we see that the spectrum of any element has cardinality $\le \dim_k(A)$.
Consider abelian C*-algebra $C^*(a)$ which is infinite dimensional ( because $\sigma(a)$ is infinite). Also $C^*(a) \subset A$ which implies that $A$ is infinite dimensional.