i was given a homework for linear algebra
and it asks:
Determine whether or not the following sets of vectors are closed under both vector addition and scalar multiplication.
how do i prove that it's closed under both conditions?
*before that, im not so sure what does it mean by "closed" under those conditions
say i have x = (x1, x2)
{x:x1 = 0}
thanks in advance
what defines a "vector space"?
2009-03-08 5:48 am
回答 (2)
2009-03-08 6:06 am
✔ 最佳答案
if you take two vectors from the set then the sum of these is also in the set , and if you multiply any vector by a scalar { non-vector} the product is a vector in the set...ex W = { <2x, 3y> | x & y are in the reals}...if you take two vectors and add { component-wise} you get a vector whose 1st entry is a multiple of 2 and 2nd entry is a multiple of 3...closed under component-wise addition, and if you multiply by any non-vector the 1st entry is still a multiple of 2 and 2nd a multiple of 3, closed under multiplication.....if R = { < even, odd> } then R is closed under addition, but if you multiply any vector by 2 you no longer have <even , odd> , thus not closed under multiplication.
2009-03-08 1:53 pm
Say v and w are any two vectors in the set. The set is closed under vector addition if v+w is always in the set.
Let c be a scalar(/number). Then the set is closed under scalar multiplication if c*v = (c v1, c v2) is always in the set.
Let c be a scalar(/number). Then the set is closed under scalar multiplication if c*v = (c v1, c v2) is always in the set.
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