Question #42ca1

Question

943 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-02T11:25:38

See below.

Calling

#X = (x_1,x_2,x_3,x_4)#
#S->((s_1),(s_2))=((1,2,1,-1),(0,-1,3,1))# and
#T-> << t_0, X >> =0# with #t_0=(1,1,0,0)#

The space #(S_|_)nn T# is

#<< s_1, X_a >> = 0#
#<< s_2,X_a >> = 0#
#<< t_0,X_a >> = 0#

This gives #X_a = lambda (-1,1,0,1), lambda in RR# so

#W# the complement, orthogonal to #X_a#. This space is given by
#X_w | << lambda(-1,1,0,1), X_w >> = 0#
This space is then

#W = << w_0, X >> = -x_1+x_2+x_4=0#

with #w_0 = (-1,1,0,1)#

As can be checked

#det((s_1),(s_2),(t_0),(w_0)) = 12#

so #{s_1,s_2,t_0,w_0}# span #RR^4#