Question

Let TεB(X,X) ,X is a comlex Banach space and T be idempotent(i.e T^2=T) .Prove that if T is neither 0 not I ,then σ(T)={0,1}

Answer 1

Question 1. Let $T \in B(X, X)$, where $X$ is a complex Banach space, and $T$ be idempotent (i.e. $T^2 = T$). Prove that if $T$ is neither 0 nor $I$, then $\sigma(T) = \{0, 1\}$.

Solution. By definition the spectrum of $T$ (denoted by $\sigma(T)$) consists of all $\lambda \in \mathbb{C}$, such that $T - \lambda I$ is not invertible.

First of all prove that $\{0,1\} \subset \sigma(T)$. If $\lambda = 0$, then $T - \lambda I = T$, so we need to show that $T$ is not invertible. Suppose there is $T^{-1}$. Then we can multiply the equation $T^2 = T$ by $T^{-1}$ and obtain $T = I$, which is not the case of our task. Now consider $\lambda = 1$ and prove that $T - \lambda I = T - I$ is not invertible. Suppose the contrary. Note that the equality $T^2 = T$ can be written as $T(T - I) = 0$. The last one, being multiplied by $(T - I)^{-1}$, implies $T = 0$, which is a contradiction.

Now take $\lambda \in \mathbb{C} \setminus \{0,1\}$ and prove that $\lambda \notin \sigma(T)$, i.e. $T - \lambda I$ is invertible. Try to find its inverse in the form $\mu T - \nu I$ for some $\mu, \nu \in \mathbb{C}$. Note that $T - \lambda I$ and $\mu T - \nu I$ commute, because $T$ and $I$ commute. So, it is sufficient to find $\mu$ and $\nu$ such that $(T - \lambda I)(\mu T - \nu I) = I$. We have

$$(T - \lambda I) (\mu T - \nu I) = \mu T ^ {2} - \nu T - \lambda \mu T + \lambda \nu I = (\mu - \nu - \lambda \mu) T + \lambda \nu I,$$

because $T^2 = T$. We need $\mu - \nu - \lambda \mu = 0$ and $\lambda \nu = 1$, so $\nu = \frac{1}{\lambda}$ and $\mu = \frac{\nu}{1 - \lambda} = \frac{1}{\lambda(1 - \lambda)}$. Since $\lambda \notin \{0,1\}$, $\mu$ and $\nu$ are defined. Thus, $\frac{1}{\lambda} \left( \frac{1}{1 - \lambda} T - I \right)$ is inverse to $T - \lambda I$ and hence $T - \lambda I$ is invertible.