compute the sprectrum of the sum of orthogonal projections

Question

Suppose $\{p_i\},i=1,\cdots,n$ are different projections and they are mutually orthogonal in a $C^*$-algebra,how to compute the spectrum $\sigma(k_1p_1+\cdots+k_np_n)$ of $ k_1p_1+\cdots+k_np_n$, where $k_1\cdots k_n>0$?

Answer 1

I will make the simplifying assumption that none of the $p_i$ are zero, and leave the removal of this assumption as an exercise.

Since $p_1,\ldots p_n$ are mutually orthogonal, the $C^*$-algebra generated by $p_1,\ldots,p_n$ and the identity, is isomorphic to $\mathbb C^{n+1}$, with an isomorphism given by $$k_1p_1+\cdots+k_np_n+k_{n+1}(1-p_1-\cdots-p_n)\mapsto(k_1,\ldots,k_n,k_{n+1})$$ In this algebra, the spectrum of $k_1p_1+\cdots+k_np_n$ is $\{0,k_1,\ldots,k_n\}$. Since this $C^*$-algebra unitally embeds into the original $C^*$-algebra, the spectrum is independent of the algebra, hence we have $$\sigma(k_1p_1+\cdots+k_np_n)=\{0,k_1,\ldots,k_n\}.$$

Or if brute force is your favorite tool, if $\lambda\notin\{0,k_1,\ldots,k_n\}$, then $$(\lambda-k_1)^{-1}p_1+\cdots+(\lambda-k_n)^{-1}p_n+\lambda^{-1}(1-p_1-\cdots-p_n)=(\lambda-k_1p_1-\cdots-k_np_n)^{-1}$$ so $\lambda\notin\sigma(k_1p_1+\cdots+k_np_n)$, and the other inclusion is easily obtained.