Spectrum of Idempotent in Algebra over Complex Numbers

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Theorem

Let $A$ be an algebra over $\C$.

Let $p \in A$ be idempotent, that is:

$p^2 = p$

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


Then:

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


Proof

Suppose first that $A$ is not unital and let $p \in A$ be idempotent.

Let $A_+$ be the unitization of $A$.

Then we have:

\(\ds \tuple {p, 0} \tuple {p, 0}\) \(=\) \(\ds \tuple {p^2 + 0p + 0 p, 0}\)
\(\ds \) \(=\) \(\ds \tuple {p, 0}\)

So $\tuple {p, 0}$ is an idempotent in $A_+$.

Since, by definition of the spectrum in a non-unital algebra, we have:

$\map {\sigma_A} p = \map {\sigma_{A_+} } {\tuple {p, 0} }$

with $A_+$ unital and $\tuple {p, 0}$ idempotent.

Without loss of generality we can therefore take $A$ to be unital.

Let $\lambda \in \C$ be such that $\lambda \not \in \set {0, 1}$.

That is, such that $\lambda \paren {1 - \lambda} \ne 0$.

We have:

\(\ds \paren {\lambda {\mathbf 1}_A - p} \paren {\paren {1 - \lambda} {\mathbf 1}_A - p}\) \(=\) \(\ds \lambda \paren {1 - \lambda} {\mathbf 1}_A - \paren {1 - \lambda} p - \lambda p + p^2\)
\(\ds \) \(=\) \(\ds \lambda \paren {1 - \lambda} {\mathbf 1}_A - p + \lambda p - \lambda p + p\)
\(\ds \) \(=\) \(\ds \lambda \paren {1 - \lambda} {\mathbf 1}_A\)

so that:

$\ds \paren {\lambda {\mathbf 1}_A - p} \paren {\frac {\paren {1 - \lambda} {\mathbf 1}_A - p} {\lambda \paren {1 - \lambda} } } = {\mathbf 1}_A$

Since $p$ and ${\mathbf 1}_A$ commute, we also have:

$\ds \paren {\frac {\paren {1 - \lambda} {\mathbf 1}_A - p} {\lambda \paren {1 - \lambda} } } \paren {\lambda {\mathbf 1}_A - p} = {\mathbf 1}_A$

Hence:

$\lambda {\mathbf 1}_A - p \in \map G A$ for $\lambda \not \in \set {0, 1}$

where $\map G A$ is the group of units of $A$.

Hence, we have:

$\map {\sigma_A} p = \set {\lambda \in \C : \lambda {\mathbf 1}_A - p \not \in \map G A} \subseteq \set {0, 1}$

$\blacksquare$


Sources

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