Polarization Identity/Complex Vector Space

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Theorem

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\C$.

Let $\norm \cdot$ be the inner product norm on $V$.


Then, we have:

$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2$

for each $x, y \in V$.


Proof

We write:

\(\ds \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2\) \(=\) \(\ds \sum_{n \mathop = 0}^3 i^n \norm {x + i^n y}^2\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^3 i^n \innerprod {x + i^n y} {x + i^n y}\) Definition of Inner Product Norm

We then compute:

\(\ds i^n \innerprod {x + i^n y} {x + i^n y}\) \(=\) \(\ds i^n \paren {\innerprod x {x + i^n y} + i^n \innerprod y {x + i^n y} }\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds i^n \innerprod x {x + i^n y} + i^{2 n} \innerprod y {x + i^n y}\)
\(\ds \) \(=\) \(\ds i^n \paren {\innerprod x x + \overline {i^n} \innerprod x y} + i^{2 n} \paren {\innerprod y x + \overline {i^n} \innerprod y y}\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds i^n \innerprod x x + i^n \paren {-i}^n \innerprod x y + i^{2 n} \innerprod y x + i^{2 n} \paren {-i}^n \innerprod y y\) Power of Complex Conjugate is Complex Conjugate of Power
\(\ds \) \(=\) \(\ds i^n \innerprod x x + \innerprod x y + \paren {-1}^n \innerprod y x + i^n \innerprod y y\)

We then have:

\(\ds \sum_{n \mathop = 0}^3 i^n\) \(=\) \(\ds 1 + i + i^2 + i^3\)
\(\ds \) \(=\) \(\ds 1 + i - 1 - i\)
\(\ds \) \(=\) \(\ds 0\)

and:

\(\ds \sum_{n \mathop = 0}^3 \paren {-1}^n\) \(=\) \(\ds \paren {-1}^0 + \paren {-1}^1 + \paren {-1}^2 + \paren {-1}^3\)
\(\ds \) \(=\) \(\ds 1 - 1 + 1 - 1\)
\(\ds \) \(=\) \(\ds 0\)

Then we have:

\(\ds \sum_{n \mathop = 0}^3 i^n \norm {x + i^n y}^2\) \(=\) \(\ds \sum_{n \mathop = 0}^3 \paren {i^n \innerprod x x + \innerprod x y + \paren {-1}^n \innerprod y x + i^n \innerprod y y}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^3 \innerprod x y + \paren {\innerprod x x + \innerprod y y} \sum_{n \mathop = 0}^3 i^n + \innerprod y x \sum_{n \mathop = 0}^3 \paren {-1}^n\)
\(\ds \) \(=\) \(\ds 4 \innerprod x y\)

$\blacksquare$


Sources

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