In the book of Linear Algebra by Greub, at page 226, it is asked that
Let $\phi$ be a linear transformation of $E$ and $\bar \phi$ the adjoint map. Denote by $|\phi|$ the norm of $\phi$ which is induced by the Euclidean norm of $E$ (cf. sec. (7.19)), i.e $$|\phi| = \sup_{||x||=1}{||\phi(x)||}.$$ Prove that $$|\phi|^2 = \lambda$$ where $\lambda$ is the largest eigenvalue of the selfadjoint mapping $\bar \phi \circ \phi$.
Note, $$(x, \phi(y)) = (\bar \phi (x), y) \quad \forall x,y \in E$$, and $E$ is a real inner product space of finite dimension.
Question:
By its definition, it is clear that
$$(\phi (x), \phi (x)) = (x, \bar \phi \phi (x)) \quad ||x|| = 1,$$ so we need to show that for a self-adjoint mapping, say $\psi$, $(x, \psi(x))$ is maximised when $x$ is a vector in eigenspace of largest eigenvalue of $\bar \psi$, but I couldn't show it, so I appreciate any help or hint about how to prove it.
As you already noticed, we need to maximize $\langle x,(\bar \phi \circ \phi) (x)\rangle$.
$\bar \phi \circ \phi$ is self adjoint, so there is an orthonormal basis of eigenvectors $v_i$ of $\bar \phi \circ \phi$, whose eigenvalues we will call $\lambda_i$ (and they are real).
Obviously, if $y$ is the normalized eigenvector with maximal eigenvalue $\lambda$, then $\langle y,(\bar \phi \circ \phi) (y)\rangle = \lambda$, so $|\phi|\ge \lambda$.
Now take an arbitrary vector $v:=\sum a_i v_i$. Then
$$\langle v,(\bar \phi \circ \phi) (v)\rangle = \sum_{i,j} \langle a_i v_i, a_j\lambda_j v_j\rangle \le \lambda\sum_{i,j} \langle a_i v_i, a_j v_j\rangle = \lambda |v|^2 .$$
As $v$ was arbitrary, we conclude $|\phi|\le \lambda$.