Well I am doing a course in functional analysis and I came across a question which I have an idea on how to solve but can't really put it on paper. I also don't know if my strategy is correct. Can someone explain why my strategy works or does not work and how I can write it out?
Question: Given $\ell^2$ with complete orthonormal system $(e_n)_n$ with $e_n = (\delta_{kn})_k$. We then define a bounded operator $T:\ell^2 \to \ell^2$. $$Te_{2k-1}=\frac{1}{k}(e_{2k-1}-ie_{2k})$$ $$Te_{2k}=\frac{1}{k}(ie_{2k-1}+e_{2k})$$
Show that $T$ is a compact self adjoint operator and calculate the spectral representation $T = \sum_{\lambda \in \sigma(T)}\lambda\pi_\lambda$. Here $\pi_\lambda$ is the projection.
My idea: Well we can give a matrix which represents $T$ with respect to $e_{2k}, e_{2k-1}$. Let's call that matrix $A$.The matrix is given by: $$A_k = \frac{1}{k}\begin{bmatrix} 1&i \\ -i&1 \end{bmatrix}$$
We can then write $T$ as a infinite matrix with blocks of $A_k$ on the diagonal. ("Jordan form") Well we define $T_n$ to be the matrix with $n$ blocks on the diagonal. This is finite rank thus if we show that $T_n \to T$, then we know that $T$ is compact. For self-adjoint I have no clue and how to calculate the spectral representation I also thought I could do something with the matrix $A_k$, but I am not sure.
Any help would be greatly appreciated!
We need to show that $T$ is compact and self-adjoint and write the operator $T$ in terms of the eigenvalues. First of all, $\pi_\lambda$ is just the orthogonal projection onto the eigenspace. More precisely, let $\{\lambda_j\}_{j = 1}^\infty$ be the set of all nonzero eigenvalues of $T$. Then $T = \sum_{j = 1}^\infty \lambda_jP_j$, where $P_j$ is the orthogonal projection onto $\mathcal{N}(T - \lambda_jI)$, i.e., $$ Tx = \sum_{j = 1}^\infty \lambda_j\langle x, u_j\rangle u_j, \ \ \textrm{ with $u_j$ eigenvector(s) associated with $\lambda_j$}. $$
Your idea of establishing compactness of $T$ by approximating it as the (operator) norm limit of a sequence of finite rank operators $T_n$ is correct so I'll let you take care of that. To show that $T$ is self-adjoint, note that since we have $T_n\to T$ in the operator norm, it suffices to show that the finite rank operators $T_n$ are bounded self-adjoint. Let us look at $Tf$ first. For any $f = (f_1, f_2, \dots)\in\ell^2$, we have $$Tf = \left(f_1 + if_2, -if_1 + f_2, \frac{1}{2}\left(f_3 + if_4\right), \frac{1}{2}\left(-if_3 + if_4\right), \dots\right). $$ One possible choice for $T_n$ for every $n\ge 1$ is the following: $$ (T_nf)_j = \begin{cases} \, (Tf)_j & \ \ \textrm{ if } \ \ 1\le j\le 2n, \\ \, 0 & \ \ \textrm{ otherwise}. \end{cases} $$ Let us now prove that $T_1$ is self-adjoint, the proof for $T_n$ is similar. \begin{align*} \langle T_1f, g\rangle & = \left(f_1 + if_2\right)\overline{g_1} + \left(-if_1 + f_2\right)\overline{g_2} \\ \langle f, T_1g\rangle & = f_1\left(\overline{g_1} - i\overline{g_2}\right) + f_2\left(i\overline{g_1} + \overline{g_2}\right) \\ & = \overline{g_1}\left(f_1 + if_2\right) + \overline{g_2}\left(-if_1 + f_2\right). \end{align*}
Finally, let's find the eigenvalues of $T$. Setting $Tf = \lambda f$, you get the following system of equations: \begin{align*} f_1 + if_2 & = \lambda f_1 \\ -if_1 + f_2 & = \lambda f_2 \\ f_3 + if_4 & = 2\lambda f_3 \\ -if_4 + f_4 & = 2\lambda f_4 \\ \vdots \qquad \vdots & = \ \ \vdots \end{align*}
Solving the first two equations, we see that either $f_1 = f_2 = 0$, or $(\lambda - 1)^2 = 1$. Solving the next two equations, we see that either $f_3 = f_4 = 0$ or $(2\lambda - 1)^2 = 1$. Continuing in this fashion, we see that for every $k\ge 1$, either $f_{2k - 1} = f_{2k} = 0$ or $(k\lambda - 1)^2 = 1$. It follows that the set of eigenvalues is $\{0\}\cup\{2/k\}_{k = 1}^\infty$. I will leave it as an exercise for you to find the eigenvectors.