Is it right to say that if two vectors, $A$ and $B$, have same $L^p$ norms, for all $p$, then $A = B$? Is it right to say that if two vectors, $A$ and $B$ (all elements of $A$ and $B$ are positive), have same $L^p$ norms, for all p, then $A = B$ ?. Thanks.
 A: As Crostul and Simon S pointed out, there is always the possibility of a permutation of the coordinates which leaves the norm invariant.
But one can show that these are indeed the only such possibilities i.e. 
that if two increasing sequences $(a_i)_{i=1}^n, (b_i)_{i=1}^n$ with positive elements satisfy
$$\sum_{i=1}^n a_i^p=\sum_{i=1}^n b_i^p$$
for all $p$ then you must have $a_i=b_i$ for all $i$.
This can be proved by an iterated argument using the fact that for $p \to \infty$ the $p$-norm of a vector assumes the value of the maximal coordinate.
Hence we conclude that $a_n=b_n$ must hold.
Then, looking at the sequences $(a_i)_{i=1}^{n-1}$ and $(b_i)_{i=1}^{n-1}$, they must still satisfy the equation and hence we conclude $a_{n-1}=b_{n-1}$
Iterating this we obtain $a_i=b_i$ for all $i$. (Writing formally, we use mathematical induction here.)
So the possibilities of permutation mentioned above are indeed the only such possibilities.
A: No. For a  vector $v=(x_1, \dots, x_n)$ the $p$-norm is $$||v||_p = \left( \sum_i |x_i|^p \right)^{\frac1p}$$
so you can see that for any permutation of indexes $\sigma \in S_n$ you have
$$||v||_p = ||(x_{\sigma(1)}, \dots, x_{\sigma(n)})||_p$$
A: Take A=(1,1,2) and B=(1,2,-1) and A is not equal to B
For p=2 , N(A)=sqrt(1^2+1^2+2^2)=sqrt(6)
          N(B)=sqrt(1^2+2^2+1^2)=sqrt(6)
  i.e N(A)=N(B)
But A is not equal to B
Therefore This result is not true in general
