ECE 130C Homework 4 Solution
May 3, 2007
2.3.14 The dimension of Sis (a) 0 when x= 0 (b) 1 when
x= (1,1,1,1) (c) 3 when x= (1,1,−1,−1) because all arrangements of
this xare perpendicular to (1,1,1,1) (d) 4 when the x’s are not equal and
don’t add to zero. No xgives dimS=2.
2.3.26 (a) True. (b) False because the basis vectors may not be in S.
2.3.34 If v1, v2, v3is a basis for V, and w1, w2, w3is a basis for W,
then these six vectors cannot be independent and some combination is zero:
Pcivi+Pdiwi= 0. This puts Pcivi=−Pdiwiin both subspaces.
2.3.36 n−r= 17 −11 = 6 is nullspace dimension; 64 −11 = 53 is
dimN(AT)
2.3.40 (a) y(x) = e2x(b) y=x(one basis vector in each case).
2.4.28
1 1
0 2
1 0
"1 0 1
1 2 0 #=
2 2 1
2 4 0
1 0 1
;r+ (n−r) = n= 3 but
2 + 2 is 4.
2.4.32 Row space = yz plane; column space = xy plane; nullspace = x
axis; left nullspace = zaxis. For I+A: Row space = column space = R3,
nullspaces contain only zero vector.
2.4.34 (a) Elimination leads to 0 = b3−b2−b1so (−1,−1,1) is in the
left nullspace. (b) Elimination leads to b3−2b1 = 0 and b4+b2−4b1= 0,
1
