Question #c5432

1 Answer
Feb 10, 2016

#S_3# is not a subspace of #RR^3#
#S_4# is a subspace of #RR^3#

Explanation:

Given a vector space #V# over a field #F# and subset #SsubeV#, we can say that #S# is a vector space over #F# with the same operations of vector addition and scalar multiplication if and only if

  • #S# is nonempty
  • #S# is closed under vector addition
  • #S# is closed under scalar multiplication

#S_3# is not a subspace of #RR^3#. While two of the conditions are fulfilled, #S_3# is not closed under vector addition. For example:
#v_1 = ((0),(1),(0)) in S_3# and #v_2 = ((1),(0),(0)) in S_3#, but #v_1+v_2=((1),(1),(0)) !in S_3#

#S_4# is a subspace of #RR^3#. To see this, we can simply check the three conditions listed above.

Nonempty:
#((0),(0),(0)) in S_4# and thus #S_4# is nonempty.

Closure under vector addition:
Let #v_1 = ((0),(0),(z_1)) in S_4# and #v_2=((0),(0),(z_2)) in S_4# be arbitrary. Then:

#v_1+v_2 = ((0),(0),(z_1))+((0),(0),(z_2))=((0),(0),(z_1+z_2)) in S_4#

Thus #S_4# is closed under vector addition.

Closure under scalar multiplication:
Let #v=((0),(0),(z)) in S_4# and #c in RR# be arbitrary. Then:

#cv = c((0),(0),(z)) = ((0),(0),(cz)) in S_4#

Thus #S_4# is closed under scalar multiplication.