Suppose $A_{n \times n}$ is a matrix and $A' = (A_{ij})$ is its entry wise absolute form, can be give an upper bound and lower bound of the L_p norm $\A\_p$ using the L_p norm of the absolute matrix $\A'\_p$.

Okay, so it's established that $\ A \_p$ means the induced norm. A few basic facts: $$ \ A \_1 = \max_{j} \sum_{i} a_{ij} \le n^{11/p} \ A \_p, $$ $$ \ A \_\infty = \max_{i} \sum_{j} a_{ij} \le n^{1/p} \ A \_p, $$ $$ \ A \_p \le \A\_1^{1/p} \ A \_\infty^{11/p}. $$ The first two lines are elementary (the inequalities following from standard comparisons of $\ell_p$ norms for vectors), and the third is a finitedimensional version of the Riesz–Thorin theorem. Putting these together, $$ \A'\_p \le \ A' \_1^{1/p} \ A' \_\infty^{11/p} = \ A \_1^{1/p} \ A \_\infty^{11/p} \le n^{\frac{2}{p}(1 \frac{1}{p})} \A\_p. $$ When $p = 2$ and $A$ is a Hadamard matrix this is sharp, and of course it's sharp for $p=1$ or $p = \infty$. I'd guess it's sharp always but I haven't thought about it. As noted by Pietro, $\ A \_p \le \ A' \_p$ always. 


There is quite a bit of information on this in Chapter 5 of Horn and Johnson "Matrix Analysis" (Cambridge University Press 1985). Perhaps it is even an exercise there. :) 


The matrix L_p norm means $\max\limits_{\x\_p = 1} \Ax\_p $, here $x$ is an $n \times 1$ vector, and the $Ax$ is an $n \times 1$ vector too. so when $p=2$, it is the well known operator norm. Thanks. I try to answer the question in a particular sense. It can be seen that even for the operator norm, that is $p=2$, when n is really large, a entry wise random $+1,1$ Bernoulli matrix have the largest singular value similar to $\sqrt{n}$, but the absolute matrix have the largest singular value $n$, so asymptotically, when $n$ is really large, we cannot have a constant $C>0$ such that $C \cdot \A \_2 \geq \A' \_2$. As @Pietro Majer has said, I think it is easy to get $\A \_p \leq \A' \_p$. but the other way around for a fixed n is still hard to me. 

