Vectors+ 06 More N-vectors: UNIZOR.COM - Math+ &Problems - Vectors

Notes to a video lecture on http://www.unizor.com

Vectors+ 06
Cauchy-Schwarz-Bunyakovsky Inequality

This inequality is also known as Cauchy-Bunyakovsky or Cauchy-Schwarz inequality.
We will consider this inequality only for a case of N-dimensional real vectors, where it states that scalar product of two vectors is less or equal to a product of their magnitudes
R·S ≤ ||R||·||S||
In terms of components,
R (R₁,R₂,...,R_N)
S (S₁,S₂,...,S_N)
R·S = R₁·S₁+R₂·S₂+...+R_N·S_N
||R|| = √R₁²+R₂²+...+R_N²
||S|| = √S₁²+S₂²+...+S_N²
To prove the above inequality, it is sufficient to prove it for squares of both left and right parts, since the right part is always positive, while the left (smaller or equal) can take negative values as well. This allows to get rid of the square roots.

Let's prove that
(R·S)² ≤ (||R||·||S||)²
In coordinate form we have to prove that
(R₁·S₁+R₂·S₂+...+R_N·S_N)² ≤
(R₁²+R₂²+...+R_N²)·(S₁²+S₂²+...+S_N²)

Proof
Consider a quadratic polynomial for an unknown variable x
P(x) = (R₁·x+S₁)²+(R₂·x+S₂)²+...
...+(R_N·x+S_N)²
This quadratic polynomial is, obviously, non-negative, as it contains only non-negative components added together.
This same polynomial can be regrouped by opening parenthesis and presented as
P(x) = (R₁²+R₂²+...+R_N²)·x² +
+ 2·(R₁·S₁+R₂·S₂+...+R_N·S_N)·x +
+ (S₁²+S₂²+...+S_N²) =
= Ax² + Bx + C
where
A = ||R||²
B = 2·R·S
C = ||S||²
In order to have a non-negative polynomial, its necessary and sufficient to have its discriminant to be non-positive.
Therefore,
B²−4·A·C ≤ 0
(2·R·S)² − 4·||R||²·||S||² ≤ 0
(R·S)² ≤ ||R||²·||S||²
End of Proof.


Consequences

1. Triangle inequality
Triangle inequality deals with a magnitude of three vectors R, S and their sum R+S that completes a triangle in N-dimensional space.
In particular, the Triangle inequality states that in any triangle sum of two sides as greater or equal than the third one.
As we know,
||R+S||² = (R+S)·(R+S) =
= R·R + S·S + 2·R·S ≤
[applying the Cauchy-Schwartz inequality to the third term]
≤ ||R||² + ||S||² +2·||R||·||S|| =
= (||R||+||S||)²
Therefore, ||R+S|| ≤ ||R||+||S|| which is the Triangle inequality.

2. The cosine between vectors
We have defined a cosine of the angle between vectors R and S in N-dimensional space as
cos(φ) = R·S/(||R||·||S||).
Because of the Cauchy-Schwartz inequality the above expression is always between −1 and 1 and, therefore, can be a cosine of some angle. So, our definition of an angle in N-dimensional space makes total sense.