Determine a base and dimension of each of the subspaces below?

#a) S={(x,y,z,w)∈ R^4|x+y-2z=0 and y+3w=0#}

#b) S={A∈M_2 (R) |tr(A)=0} and tr(A)=a_11 +a_22#

1 Answer
May 13, 2018

a) basis : #{(2,0,1,0),(3,-3,0,1)}#, dimension 2

b) basis : #{((1,0),(0,-1)),((0,1),(0,0)),((0,0),(1,0))}#, dimension 3

Explanation:

a) Since #S# is defined by imposing 2 constraints on the 4-dimensional space #RR^4#, it seems rather obvious that this is two dimensional (of course, for this argument to work, we must make sure that the two constraints are actually independent).

We can find a basis (and hence settle the question of the dimension) by directly using Gauss-Jordan elimination on the coefficient matrix. In this case, though - the system of constraints are so simple that we can directly write down their implications :

#y = -3w#

and

# x-3w-2z = 0 implies x = 2z+3w#

So, a general element of the set #S# is given by

#(2z+3w,-3w,z,w) = z(2,0,1,0)+w(3,-3,0,1)#

Thus all vectors of #S# are linear combinations of #(2,0,1,0)# and #(3,-3,0,1)# - so that the set

#{(2,0,1,0),(3,-3,0,1)}#

is a spanning set. To check whether it is linearly independent we set

#a(2,0,1,0)+b(3,-3,0,1) = (0,0,0,0)implies#

#(2a+3b,-3b,a,b) = (0,0,0,0)#

which shows that the only combination of #(2,0,1,0)# and #(3,-3,0,1)# that gives #(0,0,0,0)# is #a=0, b=0#. Thus the set

#{(2,0,1,0),(3,-3,0,1)}#

is a linearly independent spanning set - and hence a basis. So, the dimensionality of #S# is, indeed, 2.

b) A general matrix in #S# here will be

#((a_11, a_12),(a_21,a_22)) = ((a_11, a_12),(a_21,-a_11))#

where we have used the condition that the matrix is traceless. Thus we can write a general element of #S# as

#a_11((1,0),(0,-1))+a_12((0,1),(0,0))+a_21((0,0),(1,0))#

Thus

#{((1,0),(0,-1)),((0,1),(0,0)),((0,0),(1,0))}#

is a spanning set. It is easy to see that the linear combination # ((a_11, a_12),(a_21,-a_11))# is the null matrix #((0,0),(0,0))# only if #a_11 = a_12=a_21=0#. Thus the set is linearly independent as well and hence is a basis. The dimensionality in this case is 3.