Cauchy-Schwarz Inequality/Complex Numbers

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Theorem

$\ds \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$

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:

$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$

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


This gives us:

\(\ds \paren {\sum_{i \mathop = 1}^n w_i \overline {w_i} } \paren {\sum_{j \mathop = 1}^n \overline {z_j} z_j}\) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^n w_i z_i} \paren {\sum_{j \mathop = 1}^n \overline {z_j} \overline {w_j} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {w_i \overline {z_j} - w_j \overline {z_i} } \paren {\overline {w_i} z_j - \overline {w_j} z_i}\)
\(\ds \leadsto \ \ \) \(\ds \paren {\sum_{i \mathop = 1}^n w_i \overline {w_i} } \paren {\sum_{j \mathop = 1}^n \overline {z_j} z_j}\) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^n w_i z_i} \overline {\paren {\sum_{i \mathop = 1}^n w_i z_i} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {w_i \overline {z_j} - w_j \overline {z_i} } \overline {\paren {w_i \overline {z_j} - w_j \overline {z_i} } }\)
\(\ds \leadsto \ \ \) \(\ds \paren {\sum_{i \mathop = 1}^n \cmod {w_i}^2} \paren {\sum_{j \mathop = 1}^n \cmod {z_j}^2}\) \(=\) \(\ds \cmod {\sum_{i \mathop = 1}^n w_i z_i}^2 + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \cmod {w_i \overline {z_j} - w_j \overline {z_i} }^2\) Modulus in Terms of Conjugate
\(\ds \leadsto \ \ \) \(\ds \paren {\sum_{i \mathop = 1}^n \cmod {w_i}^2} \paren {\sum_{j \mathop = 1}^n \cmod {z_j}^2}\) \(\ge\) \(\ds \cmod {\sum_{i \mathop = 1}^n w_i z_i}^2\) Complex Modulus is Non-Negative

Hence the result.

$\blacksquare$


Source of Name

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


Sources

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