Let $T: \ell^2 \to \ell^2$ defined as follows: $T(x_1, x_2, x_3, x_4, x_5.., x_n,..) = (0, x_1, 0, x_4, x_5 .., x_n,..)$. Find the spectrum of $T$, the eigenvalues of $T$ (if they exist) and the corresponding eigenspaces. Is $T$ compact?
My attempt: I computed the set of eigenvalues $e(T)$ and I found that $e(T) = \left\{0,1\right\}$. Now, my claim is that $e(T) = \sigma (T)$, i.e. the eigenvalues cover the entire spectrum. To prove this, I wrote the equation $T(x) – \lambda x = b$, where $b \in \ell^2$, to show that for $\lambda \neq 0,1$ the equation is invertible in $\ell^2$. I found that $x = (x_1, x_2, x_3, x_4, …) = (\frac{-b_1}{ \lambda}, \frac{-b_2}{ \lambda} – \frac{b_1}{\lambda^2}, -\frac{b_3}{\lambda}, \frac{b_4}{(1 – \lambda)} ,…,)$, which is in $\ell^2$ because $||x||^2 = \frac{
b_1^2}{ \lambda^2} + \frac{b_2^2}{ \lambda^2} + \frac{b_1^2}{\lambda^4} + \frac{2b_1b_2}{\lambda^3} + \frac{b_3^2}{\lambda^2} + \sum_{k= 4}^{\infty} \frac{b_4^2}{(1 – \lambda)^2} = K_1 + K_2||b||^2$, where $K_1$ e $K_2$ are constants. So, $x \in \ell^2$ and $\sigma(T) = \left\{0,1\right\}$. I concluded that $T$ is not compact because otherwise the eigenvalues should have been a sequence converging to $0$. Or $T$ is not compact because the eigenspace corresponding to $1$ should have been finite dimensional, which is not.
Is this correct?
Best Answer
Your arguments are fine !
My arguments: we have $T=I-S$, where
$$S(x_1,x_2,x_3,....):=(x_1, x_2-x_1, x_3, 0,0,...).$$
It is easyto see that the set of eigenvalues of $S$ is given by $\{0,1\}.$ since $S$ is of finite rank, and hence compact, we have that $S$ has the spectrum $\sigma(S)=\{0,1\}.$
Now let let $\lambda \in \sigma(T)$. Then there is $ \mu \in \sigma(S)$ such that $\lambda =1- \mu.$ This gives $\sigma(T)=\{0,1\}.$
Suppose that $T$ is compact, Then $I=T+S$ is compact, but this is a contradiction, since $ \dim \ell^2 = \infty.$ Hence $T$ is not compact.