In analyzing a particular algorithm, we determine that it has a complexity of 
O(n). Is it possible to also describe this algorithm as O(n^2) as well
(answer yes or no explicitly)?

If you answered "no," explain why. If you answered "yes," explain why.

If possible, please answer with your given background regarding this
field of study, and any associated literature regarding the field of
complexity, big-O notation, and any related fields.

Thanks!

Not truthfully.  If it has O(n) it does not also have O(n^2), just
like a polynomial has a term of highest order.

See for example Numerical Recipies books for a readable introduction.
Various computer science and cryptogphy books also cover it.

YES 
if the running time of the program is o(n) you can say that the
running time of this algurithm is o(n^2) too
actually when you can say that the runing time of algurithm is O(n)
means that the maximun time for running is "n" times ( n is the size
of the input ) but n is the best maximum time . you can say any time
that further than "n" but "n" is the best . like 4>3 and we can say
that 7>3 and 124>3 but the best answer is 4 because 4 is the suprimum
because 4 is the minimum number that dos not further than 3 .
best regards

As a tiebreaker, yes.  The definition of big-O is that the limit of
the ratio exists (is less than infinity).  If your algorithm has a
complexity of O(n), say taking time f(n) on input of size n, this
means that lim n -> infinity of f(n)/n exists.  But then lim n ->
infinity of f(n)/n^2 also exists (it will in fact be 0).  See
http://en.wikipedia.org/wiki/Big_O_notation for more information (the
second regarding "limit superior").