University:
University of Michigan
Course:
MATH 217 | Linear Algebra
Academic year:

2024
Views:

384

Pages:

1
Author:

francescag23b

A QUICK PROOF OF THE CAUCHY-SCHWARTZ INEQUALITY Let u and v be two vectors in Rn . The Cauchy-Schwartz inequality states that |u · v| ≤ |u||v|. Written out in coordinates, this says |u1 v1 + u2 v2 + · · · + un vn | ≤ q q u21 + u22 + · · · + u2n v12 + v22 + · · · + vn2 (∗). This equation makes sure that vectors act the way we geometrically expect. For example, we know that |u + v|2 = (u + v) · (u + v) = u · u + v · v + 2u · v = |u|2 + |v|2 + 2u · v. Using Cauchy-Schwartz, we have |u|2 + |v|2 + 2u · v ≤ |u|2 + |v|2 + 2|u||v| = (|u| + |v|)2 . So the Cauchy-Schwartz inequality tells us that |u + v|2 ≤ (|u| + |v|)2 or |u + v| ≤ |u| + |v|. In other words, the length of the sum of two vectors is no more then the sum of the lengths of the vectors. As explained in class, if you believe that vectors in hundreds of dimensions act like the vectors you know and love in R2 , then the Cauchy-Schwartz inequality is a consequence of the law of cosines. Specifically, u · v = |u||v| cos θ, and cos θ ≤ 1. In case you are nervous about using geometric intuition in hundreds of dimensions, here is a direct proof. First, note that we have w · w = w12 + w22 + · · · wn2 ≥ 0 for any w. We rescale u and v to new vectors which have the same length; namely |v|u and |u|v. We take the difference of these two vectors: |u|v − |v|u. So this is a vector which vanishes if u is a positive multiple of v. We compute the dot product of this vector with itself: 0 ≤ (|u|v − |v|u) · (|u|v − |v|u) = |u|2 (v · v) − 2|u||v|(u · v) + |v|2 (u · u) = 2|u|2 |v|2 − 2|u||v|(u · v). Rearranging 2|u||v|(u · v) ≤ 2|u|2 |v|2 u · v ≤ |u||v| A similar argument using (|u|v + |v|u) · (|u|v + |v|u) shows that −u · v ≤ |u||v| as well. So |u · v| ≤ |u||v| as we promised.