The Polarization Identities

If we have an inner product on a real or complex vector space, we get a notion of length called a “norm”. It turns out that the norm completely determines the inner product.

Let’s take the sum of two vectors $v$ and $w$ . We can calculate its norm-squared as usual:

$\displaystyle\begin{aligned}\lVert v+w\rVert^2&=\langle v+w,v+w\rangle\\&=\langle v,v\rangle+\langle v,w\rangle+\langle w,v\rangle+\langle w,w\rangle\\&=\lVert v\rVert^2+\lVert w\rVert^2+\langle v,w\rangle+\overline{\langle v,w\rangle}\\&=\lVert v\rVert^2+\lVert w\rVert^2+2\Re\left(\langle v,w\rangle\right)\end{aligned}$

where $\Re(z)$ denotes the real part of the complex number $z$ . If $z$ is already a real number, it does nothing.

So we can rewrite this equation as

$\displaystyle\Re\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v+w\rVert^2-\lVert v\rVert^2-\lVert w\rVert^2\right)$

If we’re working over a real vector space, this is the inner product itself. Over a complex vector space, this only gives us the real part of the inner product. But all is not lost! We can also work out

$\displaystyle\begin{aligned}\lVert v+iw\rVert^2&=\langle v+iw,v+iw\rangle\\&=\langle v,v\rangle+\langle v,iw\rangle+\langle iw,v\rangle+\langle iw,iw\rangle\\&=\lVert v\rVert^2+\lVert iw\rVert^2+\langle v,iw\rangle+\overline{\langle v,iw\rangle}\\&=\lVert v\rVert^2+\lVert w\rVert^2+2\Re\left(i\langle v,w\rangle\right)\\&=\lVert v\rVert^2+\lVert w\rVert^2-2\Im\left(\langle v,w\rangle\right)\end{aligned}$

where $\Im(z)$ denotes the imaginary part of the complex number $z$ . The last equality holds because

$\displaystyle\Re\left(i(a+bi)\right)=\Re(ai-b)=-b=-\Im(a+bi)$

so we can write

$\displaystyle\Im\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v\rVert^2+\lVert w\rVert^2-\lVert v+iw\rVert^2\right)$

We can also write these identities out in a couple other ways. If we started with $v-w$ , we could find the identities

$\displaystyle\Re\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v\rVert^2+\lVert w\rVert^2-\lVert v-w\rVert^2\right)$
$\displaystyle\Im\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v-iw\rVert^2-\lVert v\rVert^2-\lVert w\rVert^2\right)$

Or we could combine both forms above to write

$\displaystyle\Re\left(\langle v,w\rangle\right)=\frac{1}{4}\left(\lVert v+w\rVert^2-\lVert v-w\rVert^2\right)$
$\displaystyle\Im\left(\langle v,w\rangle\right)=\frac{1}{4}\left(\lVert v-iw\rVert^2-\lVert v+iw\rVert^2\right)$

In all these ways we see that not only does an inner product on a real or complex vector space give us a norm, but the resulting norm completely determines the inner product. Different inner products necessarily give rise to different norms.

April 23, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

5 Comments »

[…] the other hand, what if we have a norm that satisfies this parallelogram law? Then we can use the polarization identities to define a unique inner […]

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[…] now we have the polarization identities to work with! The real and imaginary parts of are completely determined in terms of expressions […]

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[…] if for all vectors , then we can use the polarization identities to conclude that […]

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[…] at a given point, while the line element is a quadratic function of a single vector. However, the polarization identities will allow you to recover the bilinear function from the quadratic […]

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[…] verify our assertion for the product , we turn and recall the polarization identities from when we worked with inner products. Remember, they told us that if we know how to calculate […]

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