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.


1 Answer 1


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 $\langle x,\phi\rangle = 0$?

  • $\begingroup$ 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. $\endgroup$
    – sesodesa
    Feb 15, 2020 at 20:02
  • $\begingroup$ Or at the very least I'm going to need to provide some argument for discarding the $\lambda = 0$ case. $\endgroup$
    – sesodesa
    Feb 15, 2020 at 20:07
  • $\begingroup$ Ah yeah you're right. I was even thinking about that because it's finite rank but it just slipped my mind. Let me add to that. $\endgroup$ Feb 15, 2020 at 20:13

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