TheoremProved

Cauchy–Schwarz inequality

Statement

For real (or complex) vectors $u=(u_1,\dots,u_n)$ and $v=(v_1,\dots,v_n)$ , $\left|\sum_{i=1}^n u_i\overline{v_i}\right|^2 \le \left(\sum_{i=1}^n |u_i|^2\right)\left(\sum_{i=1}^n |v_i|^2\right)$ , with equality iff $u$ and $v$ are linearly dependent.

Why is it true?

The inequality is a disguised statement about angles: $\sum u_iv_i$ is the dot product $\lVert u\rVert\lVert v\rVert\cos\theta$ , and $\cos\theta$ can never exceed $1$ in absolute value. So the inequality just says that projecting one vector onto another can never produce something longer than the vectors themselves allow — the projection is largest exactly when the vectors point in the same (or opposite) direction.

Proof sketch

For real vectors, consider the quadratic $f(t)=\sum_i (tu_i - v_i)^2 = t^2\sum u_i^2 - 2t\sum u_iv_i + \sum v_i^2 \ge 0$ for all $t \in \mathbb{R}$ : its discriminant must be $\le 0$ , which rearranges directly into the inequality. In general (for both real and complex vectors), Lagrange's identity $\left(\sum_i |u_i|^2\right)\left(\sum_i |v_i|^2\right) - \left|\sum_i u_i\overline{v_i}\right|^2 = \sum_{i<j} |u_iv_j - u_jv_i|^2 \ge 0$ yields the inequality immediately and shows that equality holds iff $u_iv_j = u_jv_i$ for all $i,j$ , i.e. $u$ and $v$ are linearly dependent.

Proved by

proved 1821Augustin-Louis Cauchy

Topics that use this theorem

Inequalities

Related theorems

AM–GM inequality

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

G. H. Hardy, J. E. Littlewood, G. Pólya (1934). Inequalities