Determine all eigenvalues of T

Question

Let $X$ be a Hilbert space and let $T \in B(X)$ be of the form $$Tx=(x,\phi)\psi, \quad x \in X,$$ where $\phi,\psi \in X$.

The question is to determine all eigenvalues of $T$. It has the following solution:

\begin{align} Tx=\lambda x &\iff (x,\phi)\psi=\lambda x \\ &\iff \exists c \in \mathbb{K} : x =c\psi \\ &\iff \exists c \in \mathbb{K} : c(\psi,\phi)\psi = \lambda c \psi \\ &\iff (\psi,\phi)=\lambda \end{align}

I don't see why this constant $c$ would exist. Can anyone elaborate on this?

Answer 1

Let $x\neq 0$ and $x\neq\ker(T)$ (otherwise there is nothing to show) then $$Tx=\lambda x\Leftrightarrow \langle x,\phi\rangle\psi=\lambda x\Rightarrow \langle\langle x,\phi\rangle\psi,\phi\rangle=\langle\lambda x,\phi\rangle\Rightarrow\langle x,\phi\rangle\langle \psi,\phi\rangle=\lambda\langle x,\phi\rangle$$ Therefore we have $$\langle x,\phi\rangle\langle \psi,\phi\rangle-\lambda\langle x,\phi\rangle=0\Leftrightarrow \langle x,\phi\rangle(\langle \psi,\phi\rangle-\lambda)=0\Rightarrow \langle \psi,\phi\rangle=\lambda$$