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anonymous
 3 years ago
Construct a subset of XY plane R^2 that is: a) closed under vector addition & subtraction, but not under scalar multiplication. b) closed under scalar multiplication, but not vector addition & subtraction
anonymous
 3 years ago
Construct a subset of XY plane R^2 that is: a) closed under vector addition & subtraction, but not under scalar multiplication. b) closed under scalar multiplication, but not vector addition & subtraction

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is a 'trick' question in that these sets are null/empty. Strictly speaking what follows doesn't depend on it being within R^2 especially. Not even the zero vector alone qualifies. :) Scalar multiplication can be viewed as successive additions or subtractions eg. \[3V = V + V + V\]\[4V = V V V V\]and by simple extension from the integer scalars above to the rational by inversion eg. V/3 is that vector when added three times to itself equals V \[W + W + W = V \]hence\[W = V/3\]so 4V/3 is four of those \[W + W + W + W = V/3 + V/3 + V/3 + V/3 = 4V/3\]etc, and you progress to real scalars by a limiting argument ie. including the irrationals. So if you admit that a set is closed under addition AND subtraction then you must permit closure under scalar multiples too. The implications in the above argument run either way so (a) and (b) of the question are covered by the same logic. Now I think the interesting point to bring out here is that vector spaces only require closure under addition and closure under scalar multiplication. Subtraction is not mentioned. So here is, in R^2, a set for which vector addition is closed but not multiplication by an arbitrary scalar :dw:1359940364463:dwmeaning the set of vectors in R^2 with only positive components ( upper right quadrant ). The key feature to note is that the vector which is the additive inverse ( 'subtraction' ) : \[V + ( V) = 0\]is also the vector that is the scalar multiple when using minus one : \[(V) = (1)V\]it's just that by long experience with numeric algebra we may not have made that distinction earlier.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a) Consider lattice of all vectors with integer coordinates. This set is closed under vector addition&subtraction, but not under scalar multiplication, for example by 0.5. b) All vectors with coordinates (x, y) where x <=y. It is closed under scalar multiplication but not under addition , for example (1,1) +(1,1) = (0,2)
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