Linear Algebra (Math 311) – Final

Let α denote the angle between x and y, then
cosα =
〈x, y〉 1
=
||x|| · ||y|| 3 .
Let us first orthogonalize the vectors, resulting in
v1 = x and v2 = y −
〈y, x〉 1
x = y −
〈x, x〉 3 x.
The norms of these vectors are
||v1|| = √
3 and ||v2|| = 〈y − 1
1
x, y − x〉 = 3 +
3 3 1
1 2
− 2 · = 2
3 3 3 .
Normalizing the vectors v1 and v2 we find the orthonormal basis consisting
of
u1 = 1
√ x and u2 =
3 1
3
y −
2 2
1
3 x
.
Problem 8. [15 points] Show, if the columns of C are linearly independent,
then C t C is invertible.
Solution: Suppose the size of C is m × n. Let x ∈ R n . Then Cx is a
linear combination of the columns of C, and because the columns of C are
linearly independent, the only solution of Cx = 0 is the trivial solution x = 0.
Suppose C t C is singular, then there exists some non–zero vector x ∈ R n so
that C t Cx = 0. It follows that x t C t Cx = (Cx) t Cx = 0, and this means that
Cx = 0. This is a contradiction, and it follows that C t C is non–singular,
hence invertible.
6