Course 111: Algebra, 8th January 2007
To be handed in at tutorials on Jan 8th and 9th.
1. Given 3 vectors, a = (1, 2, 3), b = (4, 5, 6), c = (2, 2, 6). Show that the
following properties of the cross product hold.
• anticommutativity: a × b = −b × a = (−3, 12, −3)
• distributative over addition: a × (b + c) = (a × b) + (a × c) =
(3, 6, −5).
• scalar multiplication: (ra)×b = a×(rb) = r(a×b) = (−3r, 12r, −3r),
for a scalar r.
• Jacobi identity: a × (b × c) + b × (c × a) + c × (a × b) = 0.
• Lagrange identity: a × (b × c) = b(a · c) − c(a · b) = (32, 56, −48)
(the BAC minus CAB rule)
2. Assume that v and w are linear independent vectors. Prove that v and
v + w are linear independent vectors.
Proof: Consider rv + s(v + w) = 0 and so (r + s)v + sw = 0 since v
and w are linearly independent. Then r + s = 0 and s = 0. This means
r = s = 0 and so v and v + w are linearly independent.
3. Show that the vectors (1, 2, 3), (2, 3, 4), (3, 4, 5) do not form a basis for
R3 .
Answer If the vectors do not form a basis then they are linearly dependent and there are not scalars k, l, m not all zero such that k(1, 2, 3) +
l(2, 3, 4) + m(3, 4, 5) = (0, 0, 0). Solving the corresponding system of
equations
1k + 2l + 3m = 0
2k + 3l + 4m = 0
3k + 4l + 5m = 0
for k, l, m does not give k = l = m = 0 ie the trivial solution. Eg., one
can arrive at (−2 + 2)l = 0 which is zero for all l. Therefore the vectors
are linearly dependent and do not form a basis.
4. Find a basis of R3 containing (1, 2, 3) and (0, 1, 2).
Answers To find a basis for R3 containing (1, 2, 3) and (0, 1, 2) we just
need one more vector, (a, b, c) such that all three are linearly independent and (a, b, c) is orthogonal to (1, 2, 3) and (0, 1, 2).
Then (1, 2, 3) · (a, b, c) = 0 and (0, 1, 2) · (a, b, c) = 0 giving
a + 2b + 3c = 0
b + 2c = 0
implying a = c and b = −2c. Then a vector t(1, −2, 1) where t is
any real number satisties this criterion. So, (1, 2, 3), (0, 1, 2), (1, −2, 1)
forms a basis of R3 .