# Eigenvalues $\lambda$ on a Hilbert space $H$

Let $$H$$ be a Hilbert space and $$\phi \in H : \Vert\phi\Vert = 1$$ fixed. Define the operator $$T \in \mathcal L(H)$$ by $$\begin{equation} Tx = \langle x, \phi\rangle \phi\,. \end{equation}$$

Determine the eigenvalues and the corresponding eigenvectors and -spaces of $$T$$.

The eigenvalues can be solved from the eigenvalue equation $$\begin{equation}\label{eq:eigenequation} Tx = \lambda x \quad\Leftrightarrow\quad (T - \lambda I) x = 0 \,, \tag{1} \end{equation}$$ assuming $$T - \lambda I$$ is not injective. Also, if $$(e_n)$$ is a Schauder basis of $$H$$, after a bit of inner product manipulation $$Tx$$ can be written as $$\begin{equation} Tx = \langle x, \phi\rangle \phi = \left( \sum_n \langle x, e_n\rangle \langle e_n, \phi \rangle \right) \left( \sum_m \langle \phi, e_m\rangle e_m \right) = \left( \sum_n \langle x, e_n\rangle \langle e_n, \phi \rangle \left( \sum_m \langle \phi, e_m\rangle e_m \right)\right) \,. \end{equation}$$ Now this is where I'm supposed to be using equation \eqref{eq:eigenequation}, I believe. This is because if we write $$x$$ using its Fourier series representation, we have $$\begin{equation} \lambda x = \lambda \sum_n \langle x, e_n\rangle e_n = \sum_n \lambda\langle x, e_n\rangle e_n, \end{equation}$$ so setting $$\begin{equation} \left( \sum_n \langle x, e_n\rangle \langle e_n, \phi \rangle \left( \sum_m \langle \phi, e_m\rangle e_m \right)\right) = \sum_n \lambda\langle x, e_n\rangle e_n \end{equation}$$ or $$\begin{equation} \left( \sum_n \langle x, e_n\rangle \langle e_n, \phi \rangle \left( \sum_m \langle \phi, e_m\rangle e_m \right)\right) - \sum_n \lambda\langle x, e_n\rangle e_n = 0 \end{equation}$$ might allow us to solve for $$\lambda$$. But how the heck do I manipulate these indices so I end up with something I can use?

I tried opening the sums, but got nothing sensible.

I think you're better off working directly from the definition. Suppose $$\lambda \neq 0$$. If $$Tx = \lambda x$$, then $$\langle x, \phi\rangle \phi = \lambda x$$ which in turn means that $$x = \phi$$ and $$\lambda = \langle \phi, \phi\rangle = 1$$.
If on the other hand $$\lambda = 0$$, then $$Tx = 0$$, i.e. $$\langle x,\phi\rangle = 0$$. $$\phi$$ is a fixed unit vector in $$\mathcal{H}$$, so what vectors have the property $$\langle x,\phi\rangle = 0$$?
• I happen to know that there are multiple possible $\lambda$. A hint was given, where they ask us to consider $\lambda = 0$ and $\lambda \neq 0$ separately. So $\lambda = 1$ alone is probably not going to cut it in this case. Feb 15, 2020 at 20:02
• Or at the very least I'm going to need to provide some argument for discarding the $\lambda = 0$ case. Feb 15, 2020 at 20:07