Cauchy-Schwarz Inequality/Complex Numbers

Theorem

$\displaystyle \left({\sum \left\vert{w_i}\right\vert^2}\right) \left({\sum \left\vert{z_i}\right\vert^2}\right) \ge \left\vert{\sum w_i z_i}\right\vert^2$

where all of $w_i, z_i \in \C$.

Proof

Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary complex numbers.

Take the Binet-Cauchy Identity:

$\displaystyle \left({\sum_{i=1}^n a_i c_i}\right) \left({\sum_{j=1}^n b_j d_j}\right) = \left({\sum_{i=1}^n a_i d_i}\right) \left({\sum_{j=1}^n b_j c_j}\right) + \sum_{1 \le i < j \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$

and set $a_i = w_i, b_j = \overline {z_j}, c_i = \overline {w_i}, d_j = z_j $.

This gives us:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\sum_{i=1}^n w_i \overline {w_i} }\right) \left({\sum_{j=1}^n \overline {z_j} z_j}\right)\)	\(=\)	\(\displaystyle \left({\sum_{i=1}^n w_i z_i}\right) \left({\sum_{j=1}^n \overline {z_j} \overline {w_j} }\right) + \sum_{1 \le i < j \le n} \left({w_i \overline {z_j} - w_j \overline {z_i} }\right) \left({\overline {w_i} z_j - \overline {w_j} z_i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({\sum_{i=1}^n w_i \overline {w_i} }\right) \left({\sum_{j=1}^n \overline {z_j} z_j}\right)\)	\(=\)	\(\displaystyle \left({\sum_{i=1}^n w_i z_i}\right) \overline {\left({\sum_{i=1}^n w_i z_i}\right)} + \sum_{1 \le i < j \le n} \left({w_i \overline {z_j} - w_j \overline {z_i} }\right) \overline {\left({w_i \overline {z_j} - w_j \overline {z_i} }\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({\sum_{i=1}^n \left\vert{w_i}\right\vert^2}\right) \left({\sum_{j=1}^n \left\vert{z_j}\right\vert^2}\right)\)	\(=\)	\(\displaystyle \left\vert{\sum_{i=1}^n w_i z_i}\right\vert^2 + \sum_{1 \le i < j \le n} \left\vert{w_i \overline {z_j} - w_j \overline {z_i} }\right\vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Modulus in Terms of Conjugate
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({\sum_{i=1}^n \left\vert{w_i}\right\vert^2}\right) \left({\sum_{j=1}^n \left\vert{z_j}\right\vert^2}\right)\)	\(\ge\)	\(\displaystyle \left\vert{\sum_{i=1}^n w_i z_i}\right\vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\left\vert{w_i \overline {z_j} - w_j \overline {z_i} }\right\vert^2$ can not be negative

Hence the result.

$\blacksquare$

Source of Name

This entry was named for Augustin Louis Cauchy and Hermann Amandus Schwarz.

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.3$: Exercise $30$