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 $T$.
A totally incomplete answer
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.
Best Answer
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 $\langle x,\phi\rangle = 0$?