I am asked to find the spectrum of a unitary $T : \ell^2(N) \to \ell^2(N)$ (very explicitly) but as it happens, I can write in in block diagonal form. So I find square matrices $A_1, A_2, \ldots$ ($A_k$ has dimensions $n_k \times n_k$) so that if we write $x = (x_1, x_2, \ldots) = (x^{(1)}, x^{(2)}, \ldots)$ such that the length of $x^{(k)}$ corresponds to the dimension of $A_k$, then $Tx(j)= A_k x^{(K)} (i)$ where $K$ and $i$ are such the indexes correspond. I mean that we can understand $T$ as the map $$x = \begin{bmatrix} x^{(1)} \\ x^{(2)} \\ x^{(3)} \\ \vdots \end{bmatrix} \in \ell^2(\mathbb{N}) \mapsto \begin{bmatrix} A_1 & & & \\ & A_2 & & \\ & & A_3 & \\ & & & \ddots \end{bmatrix} \begin{bmatrix}x^{(1)} \\ x^{(2)} \\ x^{(3)} \\ \vdots \end{bmatrix}.$$
What can we say about the spectrum of $T$ in terms of the eigenvalues of the finite matrices $A_k$? I have been able to find that the eigenvalues of $T$ are exactly the union of the eigenvalues of the $A_k$ ($\ast$). This is not very hard. But I would like to prove that the continuous and residual spectra of $T$ are empty, so Spec $T = \{ \lambda \in C \mid \lambda \text{ is an eigenvector of some } A_k \} $. Any help is greatly appreciated!!
For completeness, let me prove ($\ast$). Let $\lambda \in \mathbb{C}$. Then $\lambda$ is an eigenvalue of $T$ iff we find nonzero $x \in \ell^2(\mathbb{N})$ such that $Tx = \lambda x$. Also, $\lambda$ is an eigenvalue of some $A_k$ iff we find $k \geq 1$ and a nonzero $n_k$ dim vector $v$ such that $A_k v = \lambda v$.
So: if $\lambda$ is an eigenvalue of $T$, then we find some nonzero $x \in \ell^2(\mathbb{N})$ such that $Tx = \lambda x$, and if we take $k \geq 1$ such that $x^{(k)}$ is a nonzero $n_k$ dim vector, then it's easily seen that $A_k v = \lambda v$. So then $\lambda$ is an eigenvalue of $A_k$.
If conversely $\lambda$ is an eigenvalue of some $A_k$, then we find a vector $v$ such that $A_k v = \lambda v$. Then, we may define $x \in \ell^2(\mathbb{N})$ by putting $x^{(j)} = 0$ for all $j \neq k$ and $x^{(k)} = v$. Then it is not very hard to see that $Tx = [0, \ldots, 0, A_k x^{(k)}, 0, \ldots]^t = [0, \ldots, 0, \lambda x^{(k)}, 0, \ldots]^t = \lambda x.$ So $\lambda$ is an eigenvalue of $T$.
The residual spectrum is empty for any unitary; more generally, for any normal operator. Because if $T$ is normal then $$ \ker T=\ker T^*T=\ker TT^*=\ker T^*. $$ Taking orthogonals we get that $\def\ran{\operatorname{ran}}\overline{\ran T}=\overline{\ran T^*}$. If $\lambda$ were in the residual spectrum of $T$, then $$\{0\}\ne(\ran(T-\lambda))^\perp=\ker(T-\lambda)^*=\ker(T-\lambda) $$ and $\lambda$ would be an eigenvalue, a contradiction.
Since the spectrum is closed, you may have elements in the spectrum of your unitary that are not eigenvalues. So the continuous spectrum can be nonempty. For example let $\{q_n\}$ be an enumeration of $\mathbb Q\cap(0,1)$. Consider a diagonal operator with diagonal $\{e^{2\pi i q_n}\}$. Since the elements of the diagonal are dense in $\mathbb T$, the continuous spectrum is $\mathbb T\setminus \{e^{2\pi i q_n}\}$, an uncountable set dense in $\mathbb T$.