Let #I# is identity matrix sized #3xx3# and #J# matrix sized #3xx3# which all the entry is 1. Let #A# is matrix sized #6xx6# which is wrote in block matrix #A=((I,J),(0,0))#. How to determine the base of zero space of #A# ?

1 Answer
Narad T.
Oct 25, 2017

See the explanation below

Explanation:

The matrices are

#I=((1,0,0),(0,1,0),(0,0,1))#

#J=((1,1,1),(1,1,1),(1,1,1))#

#A=((1,0,0,1,1,1),(0,1,0,1,1,1),(0,0,1,1,1,1),(0,0,0,0,0,0),(0,0,0,0,0,0),(0,0,0,0,0,0))#

To determine the nul space of #A#, solve #Ax=0#

#A=((1,0,0,1,1,1),(0,1,0,1,1,1),(0,0,1,1,1,1),(0,0,0,0,0,0),(0,0,0,0,0,0),(0,0,0,0,0,0))((x_1),(x_2),(x_3),(x_4),(x_5),(x_6))=((0),(0),(0),(0),(0),(0))#

#x_1=-s-t-u#

#x_2=-s-t-u#

#x_3=-s-t-u#

#x_4=s#

#x_5=t#

#x_6=u#

#((x_1),(x_2),(x_3),(x_4),(x_5),(x_6))=((-s-t-u),(-s-t-u),(-s-t-u),(s),(t),(u))#

#=s((-1),(-1),(-1),(1),(0),(0))+t((-1),(-1),(-1),(0),(1),(0))+u((-1),(-1),(-1),(0),(0),(1))#

#=svecv_1+tvecv_2+uvecv_3#

The Null Space #(A)={svecv_1+tvecv_2+uvecv_3 | s,t,u in RR}#

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