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Question 2. In question 4 of Problem Sheet 1, we saw that the set U={(x, y, z, t) E R4 | x + y + z + t = 0} C R4 is a vector space in its own right. Show that the three vectors 0 0 lie in U, are linearly independent, and span U. Hence deduce that the dimension of U is 3. The moral of this question: by imposing the single linear condition x + y+ z + t 0 on the space R4, ue get a space of dimension 3-4-1.Explanation / Answer
2. U = {(x,y,z,t)T R4| x+y+z+t = 0}. For the vector u1 = (1,0,0,-1)T, since 1+0+0+(-1) = 0, hence u1 = (1,0,0,-1)T U. Also, for the vector u2 = (0,1,0,-1)T, since 0+1+0+(-1) = 0, hence u2 = (0,1,0,-1)T U and also , for. the vector u2 = (0,0,1,-1)T, since 0+0+1+(-1) = 0, hence u3 = (0,0,1,-1)T U. Now, let A be the 4x3 matrix with u1,u2,u3 as columns. Then A =
1
0
0
0
1
0
0
0
1
-1
-1
-1
The RREF of A is
1
0
0
0
1
0
0
0
1
0
0
0
Hence u1,u2,and u3 are linearly independent. The linear independence of u1,u2,u3 can be proved even otherwise as only u1 has 1 in the 1st place, only u2 has 1 in the 2nd place and only u3 has 1 in the 3rd place so that none of these 3 vectors can be expressed as a linear combination of others. Further, an arbitrary vector in U is of the form (x,y, z, -x-y-z)T = x(1,0,0,-1)T+y(0,1,0,-1)T+z(0,0,1,-1)T = xu1+yu2+zu3. Hence U = span { u1,u2,u3 } and the dimension of U is 3.
1
0
0
0
1
0
0
0
1
-1
-1
-1
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