I'm working on the following question: Let $X$ be a Hilbert space with dim($X$)≥2 and let $T\in B(X)$ be of the form $$Tx=\langle x,\varphi\rangle\psi$$ Show that $||T||=||\varphi|| \hspace{0.3em}||\psi||$
I started in the following way: $$||Tx||^2=\langle \langle x,\varphi\rangle\psi, \langle x,\varphi\rangle\psi \rangle$$ $$=\langle x,\varphi \rangle \langle \psi,\langle x, \varphi \rangle\psi \rangle$$ $$=\langle x,\varphi \rangle \overline{\langle x,\varphi \rangle}\langle \psi,\psi \rangle$$
In the solutions I find that $$||Tx||=|\langle x,\varphi \rangle| \hspace{0.3em} ||\psi||$$
Where does the $|\langle x,\varphi \rangle| $ come from?
In your lines, you get to $\|Tx\|^2 = \langle x, \varphi \rangle\overline{\langle x, \varphi \rangle}\langle \psi, \psi \rangle$. Notice that for any complex number $z$ we have $z\overline z = |z|^2$, and for any vector $v$ in inner product space we have $\langle v, v \rangle = \|v\|^2$ (this is how norm is induced on inner product space), so together we get $$\|Tx\|^2 = |\langle x, \varphi \rangle|^2\|\psi\|^2 \implies \|Tx\| = |\langle x, \varphi \rangle|\|\psi\|.$$