Page 92 - 35Linear Algebra

92 Vectors in Space, n-Vectors

Reading homework: problem 1

Example 53 Consider a four-dimensional space, with a special direction which we will
4
2 2
1 1
call “time”. The Lorentzian inner product on R is given by hu, vi = u v + u v +
3 3
4 4
u v − u v . This is of central importance in Einstein’s theory of special relativity.
Note, in particular, that it is not positive definite. As a result, the “squared-length”
2
2
2
2
2
of a vector with coordinates x, y, z and t is kvk = x + y + z − t . Notice that
2
it is possible for kvk ≤ 0 even with non-vanishing v! The physical interpretation of
this inner product depends on the sign of the inner product; two space time points
X 1 := (x 1 , y 1 , z 1 , t 1 ), X 2 := (x 2 , y 2 , z 2 , t 2 ) are
p
• separated by a distance hX 1 , X 2 i if hX 1 , X 2 i ≥ 0.
p
• separated by a time −hX 1 , X 2 i if hX 1 , X 2 i ≤ 0.
In particular, the difference in time coordinates t 2 − t 1 is not the time between the
two points! (Compare this to using polar coordinates for which the distance between
two points (r, θ 1 ) and (r, θ 2 ) is not θ 2 − θ 1 ; coordinate differences are not necessarily
distances.)
Theorem 4.3.1 (Cauchy-Schwarz Inequality). For any non-zero vectors u
and v with an inner-product h , i

|hu, vi|
≤ 1.
kuk kvk
The easiest proof would use the definition of the angle between two vectors
and the fact that cos θ ≤ 1. However, strictly speaking speaking we did
not check our assumption that we could apply the Law of Cosines to the
n
Euclidean length in R . There is, however a simple algebraic proof.
Proof. Let α be any real number and consider the following positive, quadratic
polynomial in α

2
0 ≤ hu + αv, u + αvi = hu, ui + 2αhu, vi + α hv, vi .
2
b
Since any quadratic aα +2bα+c takes its minimal value c− b 2 when α = − ,
a 2a
and the inequality should hold for even this minimum value of the polynomial
hu, vi 2 |hu, vi|
0 ≤ hu, ui − ⇔ ≤ 1.
hv, vi kuk kvk

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