Spectrum of Projection in *-Algebra

Theorem

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.

Let $p$ be a projection on $A$.

Let $\map {\sigma_A} p$ be the spectrum of $p$ in $A$.

Then $\map {\sigma_A} p \subseteq \set {0, 1}$.

Corollary

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.

Let $p$ be a projection on $A$.

Let $\map {\sigma_A} p$ be the spectrum of $p$ in $A$.

Then:

$p = {\mathbf 0}_A$ if and only if $\map {\sigma_A} p = \set 0$

$p = {\mathbf 1}_A$ if and only if $\map {\sigma_A} p = \set 1$

$p \not \in \set { {\mathbf 0}_A, {\mathbf 1}_A}$ if and only if $\map {\sigma_A} p = \set {0, 1}$.

Proof

From the definition of a projection, we have $p^2 = p$.

Hence $p^2 - p = {\mathbf 0}_A$.

Hence from Spectrum of Zero Vector in Algebra, we have $\map {\sigma_A} {p^2 - p} = \map {\sigma_A} { {\mathbf 0}_A} = \set 0$.

From the Spectral Mapping Theorem for Polynomials, we have:

$\set 0 = \set {z^2 - z : z \in \map {\sigma_A} p}$

In other words:

if $z \in \map {\sigma_A} p$, then $z^2 - z = 0$.

That is:

$z = 0$ or $z = 1$.

So we obtain $\map {\sigma_A} p \subseteq \set {0, 1}$.

$\blacksquare$