Operator norm is equal to max eigenvalue Take a matrix $A \in M_{2 \times 2}(\mathbb{R})$ and consider the norm $\vert\vert A\vert\vert = \sup\limits_{x \in \mathbb{R}^2} \frac{ \vert\vert Ax\vert\vert}{\vert\vert x \vert\vert} = \sup\limits_{x, \vert\vert x \vert\vert = 1} \vert\vert Ax\vert\vert$. 
I am unable to see that the norm must be less than or equal to the maximum eigenvalue of $A$:
$\vert\vert A\vert\vert \le \max\limits_{\lambda \in \sigma(A)} \lambda$ 
and I am also unable of characterizing the type of $A$ such that:
$\vert\vert A\vert\vert = \max\limits_{\lambda \in \sigma(A)} \lambda$ 
 A: It is not true that $||A|| \leq \max \{|\lambda|\}$ in general. The reverse inequality always holds: $Ax=\lambda x,x \neq 0$ implies $|\lambda| ||x|| \leq ||A|| ||x||$ so $|\lambda| \leq ||A||$ for any eigen value $\lambda$. Equality holds if $A$ is symmetric.
A: *

*If $x$ is a unit eigenvector corresponding to eigenvalue $\lambda$, then $\|Ax\| = |\lambda|$, so $|\lambda| \le \|A\|$. So the inequality that you stated should be reversed.

*If $A$ is diagonalizable with respect to an orthonormal basis $\{v_1, v_2\}$ with eigenvalues $\lambda_1$ and $\lambda_2$, then for any $x=c_1 v_1 + c_2 v_2$ we have $Ax = c_1 \lambda_1 v_1 + c_2 \lambda_2 v_2$ so $\frac{\|Ax\|^2}{\|x\|^2} = \frac{c_1^2 \lambda_1^2 + c_2 \lambda_2^2}{c_1^2 + c_2^2} \le \max\{\lambda_1^2, \lambda_2^2\}$, with equality if $x$ is an eigenvector corresponding to the largest eigenvalue (in absolute value). This situation happens whenever $A$ is symmetric, or more generally, normal ($A^\top A = A A^\top$).
An example for strict inequality is if $A$ is similar to a matrix of the form $\begin{bmatrix} 0 & 1 \\ & 0\end{bmatrix}$, in which case the eigenvalues are all zero, but the operator norm is $1$.
These facts can be generalized to arbitrarily sized square matrices as well.
A: Observe we have that
\begin{align}
\|Ax\|_2^2= x^TA^TAx
\end{align}
where $A^TA$ is diagonalizable since it's symmetric. Then we see that
\begin{align}
x^TA^TAx = y^TDy
\end{align}
where $D$ consists of the square of the singular values. Hence, we get that
\begin{align}
\|Ax\|_2^2 = \lambda_1^2y_1^2+\lambda_2^2y_2^2.
\end{align}
In the case when $\|x\|_2=1$ then we also have that $\|y\|_2=1$. Hence it follows
\begin{align}
\|Ax\|_2^2 \leq \max_{1\le i \le 2}\lambda_i^2 \ \ \implies \ \ \|Ax\|_2 \leq \max_{1 \le i \le 2}\lambda_i.
\end{align}
Note that I say singular value, not eigenvalue.
Additional: Consider
\begin{align}
A=
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}
\end{align}
then we see that 
\begin{align}
\sup_{\|x\|_2=1}\|Ax\|_2 = 1
\end{align}
but the eigenvalues of $A$ are $0$.
Moreover, observe that
\begin{align}
A^TA=
\begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}
\end{align}
which means the singular values of $A$ are equal to $1$ and $0$.
