Show that $\Vert T\Vert=\max\limits_{1\leq i\leq n}\vert d_{i}\vert $ Please give me a hint for the following question. 

The Problem:
(a)  Let $D\in M_{nn}(\mathbb{R})$ be a diagonal matrix and $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be the linear operator associated with $D,$ $Tx=Dx$ for all $x\in\mathbb{R}^{n}.$ Show that 
  $$\Vert T\Vert=\max\limits_{1\leq i\leq n}\vert d_{i}\vert $$
  where $d_{1}, d_{2}, \cdots, d_{n}$ are the entries on the diagonal of $D$. 
   For this part of the question I have proved that  $\Vert T\Vert\leq\max\limits_{1\leq i\leq n}\vert d_{i}\vert $

For this part of the question I have proved that  $\Vert T\Vert\leq\max\limits_{1\leq i\leq n}\vert d_{i}\vert $.

(b)
  Suppose $A\in M_{nn}(\mathbb{R}^{n})$ is symetric (i.e.,$A^{T}=A$) and $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is the linear operator associated with $A$. Show that 
  $$\Vert T\Vert=\max\{ \vert \lambda\vert; \lambda\; is\; the\; eigenvalue\; of\; A\}. $$

 A: Hints: For (a): If you have already proven that the operator norm of $T$ is at most the maximal diagonal entry, now just recognize that there is for sure a vector $v \in \mathbb{R}^n$ which satisfies $|T v|=(max|d_i|) |v|$. This will be a vector of the form $(0,0,0,1,0,0)$ with the $1$ in the component with the largest eigenvalue.
For (b): You know that a symmetric matrix can be diagonalized, so you should try to reduce this to the statement of (a). Write $A = SDS^{-1}$ with $D$ a diagonal matrix.
A: A Rayleigh quotient approach:
Assuming $$\|T\|=\max_{x\neq 0}\frac{\|Tx\|_2}{\|x\|_2},$$
and $T$ is symmetric, then, note that 
$$R(x)=\frac{\|Tx\|_2^2}{\|x\|_2^2}=\frac{\langle Tx,Tx\rangle}{\langle x,x\rangle}=\frac{\langle T^2x,x\rangle}{\langle x,x\rangle}$$
which is nothing else than the Rayleigh quotient of $T^2$. Hence, we have 
\begin{align*}
\|T\|^2&=\max_{x\neq 0} R(x) \\
&=\max\{|\tilde \lambda|\colon \tilde\lambda \text{ is an eigenvalue of }T^2\}\\
&=\max\{|\lambda|^2\colon \lambda \text{ is an eigenvalue of }T\}\\
&=\max\{|\lambda|\colon \lambda \text{ is an eigenvalue of }T\}^2.
\end{align*}
Of course, any diagonal matrix is symmetric so the first part follows from this argument.
