Can the state of a system after applying the operation "absolute value" be got back using elementary operations or transformations? Take the operation or transformation "addition". You can get back the original state of the system by doing the opposite operation, i.e., "subtraction".
But, if the operation is "absolute value", you can't get back the original state. Same with the operation "log" - take the "antilog" and you get back your original state.
For ex, if I add 5 to every element in a vector, I just have to subtract 5 from each element of the vector to get the original vector back. But if I take absolute value of all elements of a vector, I can't get back the original vector by doing any elementary operation. Or can I?
P.S. Are there any specific terms to describe these 2 kinds of operations? Something like, tractable and untractable, for the former and the latter types, respectively, perhaps? And what are some other operations of the latter kind?
EDIT: One more interesting thing about this operation is, you can't even tell which are the elements that the operation impacted and which aren't! Even if only 1 of the entries in the vector was impacted by the operation, it caused us to be unsure about the all remaining entries in the vector too! Are there ways to at least know which elements were affected?
 A: You're looking at an injectivity problem : the function $x\mapsto |x|$ is not injective, meaning that you can have $x\neq y$ but $|x| = |y|$, so knowing only $|x|$ you can't recover $x$ (since you can't know if you started from $x$ or $-x$).
In general if you have a function $f:X\to Y$, the fact that you can recover $x$ from $f(x)$ is exactly equivalent to $f$ being injective. Of course that doesn't mean that there is an effective way to recover it, but at least it's mathematically possible.
Specifically here it means that there is a function $g: Y\to X$ such that $g\circ f = Id_X$, so that if you know $y=f(x)$, you can recover $x$ by $x = g(y)$. And the existence of such a function $g$ is equivalent to $f$ being injective. (But the function $g$ may be complicated, or impossible to compute.)
A: You can out of any vector $\in {\mathbb R}^n$ construct another vector $\in {\mathbb R}^{n+1}$ which is identical but with one added entry, that in the binary number system stores 1 for + and 0 for - and where the (base 2) exponent of the digit encodes index in the vector. 
So really it depends in some sense on a measure of the "redundancy" of the vector.
