show norm of self-adjoint operator is maximum of abs value of eigenvalue $M: V \to V$ linear operator
show
$\|M\| = \max\{|\text{eigenvalue}|\}$
 A: Recall that any self-adjoint matrix is diagonalizable in some orthonormal basis, i.e. there exists an orthonormal basis $e_1,...,e_n$ of $\mathbb{C}^n$ such that $Ae_i=\lambda_ie_i$ for $i=1,2,...,n$. Denote the linear transformation associated with matrix $A$ by $T$. Let $\lambda=\text{max} \, \{|\lambda_1|,...,|\lambda_2|\}$. Then for any $x \in \mathbb{C}^n$ such that $\|x\| \leq 1$, $x=a_1e_1+...+a_ne_n$ for some $a_1,...,a_n \in \mathbb{C}$, and thus we get that
\begin{equation}
\begin{split}
\|Tx\|
 & =\|T(a_1e_1+...+a_ne_n)\|\\
 & =\|a_1 T(e_1)+...+a_nT(e_n)\|\\
 & =\|a_1 \lambda_1e_1+...+a_n\lambda_ne_n\|\\
 & = \sqrt{|a_1 \lambda_1|^2+...+|a_n\lambda_n|^2}\\
 & \leq \sqrt{|a_1 \lambda|^2+...+|a_n\lambda|^2}\\
 & = \lambda \sqrt{|a_1|^2+...+|a_n|^2}\\
 & = \lambda \|x\|\\
 & \leq \lambda\\
\end{split}
\end{equation}
Since $x$ was an arbitrary element in $\mathbb{C}^n$ such that $\|x\| \leq 1$, we conclude that
$$\|T\|=\sup_{\|x\| \leq 1} \|Tx\| \leq \lambda$$
Conversely, notice that $|\lambda_i|=\lambda$ for some $i$. Then, for that $i$, we have
\begin{equation}
\begin{split}
\|Te_i\|
 & =\|\lambda_i e_1\|\\
 & =|\lambda_i|\|e_i\|\\
 & =\lambda\\
\end{split}
\end{equation}
Since $\|e_i\|=1$, this implies that
$$\|T\|=\sup_{\|x\| \leq 1} \|Tx\| \geq \lambda$$
Thus, we conclude that
$$\|T\|=\sup_{\|x\| \leq 1} \|Tx\| = \lambda$$
A: The norm of an operator (as derived from the norm on the space) is given by
$$
\|M\| = \max_{x \neq 0} \frac{\|Mx\|}{\|x\|} = \max_{\|x\| = 1} \|M x\|
$$
If you're referring to the Euclidean norm (i.e. the 2-norm), then
$$
\|M\| = \max_{\|x\| = 1} \sqrt{(Mx)^*Mx} = \sqrt{\max_{\|x\| = 1} x^*M^*Mx}
$$
Hint: note that we can write $M = UDU^*$ for real diagonal matrix $D$ and unitary matrix $U$.  It may help to set $y = U^*x$, noting that $\|y\| = \|x\|$.
A: This follows from the Min-max theorem, where the subspace of the Dirichlet form is the whole space.
More detail:
If $M$ is self adjoint, there is an orthonormal basis consisting of evectors of $M$, say $e_1,\ldots,e_n$, (let $\lambda_i$ be the evalues arranged so that $|\lambda_1| \geq \cdots |\lambda_n|$). For $v \in V$ let $a^v_1,\ldots,a_n^v$ be the coefficients satisfying 
$v = \sum_{i=1}^n a_i^v e_i$. Then
\begin{align}
   \|M \|^2
&= \sup\{ \langle Mv,Mv\rangle \mid \|v\| = 1 \} \\
&= \sup\left\{ \left\langle M\sum_{i=1}^n a^v_i e_i,
   M \sum_{i=1}^n a^v_i e_i\right\rangle \mid \|v\| = 1 \right\} \\
&= \sup\left\{ \left\langle \sum_{i=1}^n a^v_i \lambda_i e_i,
    \sum_{i=1}^n a^v_i \lambda_i e_i\right\rangle \mid \|v\| = 1 \right\} \\
&= \sup\left\{ \sum_{i=1}^n |a_i^v|^2 |\lambda_i|^2 \langle  e_i,
         e_i\rangle \mid \|v\| = 1 \right\} \\
&= \sup\left\{ \sum_{i=1}^n |a_i^v|^2|\lambda_i|^2 \mid \|v\| = 1 \right\} 
\end{align}
(the cross terms were $0$ because of orthogonality). Now, you have a sum of squares, and the coefficients $|a_i^v|^2$ must sum to $1$ since $\|v\|=1$. The max occurs when all mass is put at the largest component, $|\lambda_1|^2$. Therefore
$$
  \|M\|^2
= |\lambda_1|^2.
$$
