Spectral radius is not matrix norm. I have seen an example of matrix 
$$A = \begin{bmatrix}
   0 & 1  \\
   0 & 0
  \end{bmatrix}$$
whose spectral radius is zero therefore the spectral radius is not matrix norm. Why the spectral radius is not matrix norm in this case Is it possible that $\|A\|=\epsilon$?
 A: From this Wikipedia page, the spectral norm of a matrix $A\in\mathbb{C}^{n\times n}$ is defined as
$$\rho\left(A\right)=\max_{1\leq i\leq n}\left\{\left|\lambda_{i}\right|\right\}$$
where the $\lambda_{i}$'s are the eigenvalues of the matrix. In your case
$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
is triangular, so the diagonal entries are the eigenvalues. Thus you have $\lambda_{1}=\lambda_{2}=0$, and it follows immediately that
$$\rho\left(A\right)=0$$
This is not a norm since $A\neq 0$ and a norm $\left\|\cdot\right\|$ must satisfy

$$\left\|A\right\|=0\:\Longleftrightarrow\: A=0$$

by definition.
A: You seem to have confused spectral radius with spectral norm. The word spectral in spectral norm actually refers not to the spectrum of $A$, but to the spectrum of $A^\ast A$. In my opinion, we should all abandon this misleading name, but adopt the more appropriate name, the operator norm, instead.
The operator norm $\|A\|_2$ is defined by
$$
\max_{\|x\|_2\ne0}\frac{\|Ax\|_2}{\|x\|_2}=\max_{\|u\|_2=1}\|Au\|_2.\tag{1}
$$
It is identical to the largest singular value, i.e. the square root of the largest eigenvalue of $A^\ast A$.
Unlike spectral radius, the operator norm is defined for all square or non-square complex matrices. The norm can be more accurately called the spectral norm when $A$ is square and self-adjoint. In that case, the operator norm coincides with the spectral radius, and the spectral radius is a (submultiplicative) norm on the real linear space of all self-adjoint matrices.
The operator norm is always bounded below by the spectral radius, but as your example shows, the two quantities can be unequal. Your $A$ is nonzero in the first place, so its norm cannot possibly be zero. Using the equivalent definition on the RHS of $(1)$, or the singular value of $A$, it is not hard to show that $\|A\|_2=1$.
Square matrices whose operator norms equal their spectral radii are called radial. The name originates from the fact that the equality of $\|A\|_2$ and $\rho(A)$ is equivalent to the equality of $\|A\|_2$ and $r(A)=\max_{\|u\|=1}|\langle Au,u\rangle|$, the numerical radius of $A$. Radial matrices includes normal matrices as subclass, while the latter includes self-adjoint matrices as an even smaller subclass.
A: Define the norm $\| A \|_L = \|V_L^{-1} A V_L\|$
where
$V_L = \begin{bmatrix} 1 & 0 \\ 0 & {1 \over L}\end{bmatrix}$.
Note that $\lim_{L \to \infty} V_L^{-1} A V_L = 0$ 
Addendum:
Note that the above norm is an induced norm. If we let
$\|x\|_L = \|V_L^{-1} x\|$ then
$\|A\|_L = \sup_{\|x\|_L \le 1} \|Ax\|_L$.
Another addendum (the question changed):
With $U_L=\begin{bmatrix} L & 0 \\ 0 & 1\end{bmatrix}$ show
that $\lim_{L \downarrow 0} \|U_L^{-1} A U_L\| = 0$ and
$\lim_{L \to \infty} \|U_L^{-1} A U_L\| = \infty$.
Now, for any $r>0$ use the intermediate value theorem to choose some $L$ such that 
$\|U_L^{-1} A U_L \| = r$.
