Prove Polarization Formula in Complex Vector Space

[HernanV]

hi, all

Let V be a Complex Vector Space:

probe that:

$$<u\mid v> = \frac {1} {4} (\parallel u + v \parallel ^ 2 - \parallel u - v \parallel ^ 2) - \frac {\imath} {4} (\parallel u + \imath v\parallel ^ 2 - \parallel u - \imath v\parallel ^ 2) \forall u,v$$

Polarization formula.

i've multiplied both sides by 4, then aplicated internal product properties and obtained...

$$4 <u\mid v> = 4 <u\mid v> - \imath 4 u$$

please help!

thank you

 

[mathwonk]

perhaps the minus sign in front of the i/4 should be a plus sign.

 

[M98Ranger]

for your question.

To prove the polarization formula in complex vector space, we will start by defining the inner product of two vectors u and v in a complex vector space V as:

<u|v> = u*v + u*v

Where u*v represents the complex conjugate of u multiplied by v. This definition is consistent with the properties of an inner product, such as linearity and conjugate symmetry.

Now, let's consider the left-hand side of the polarization formula:

<u|v> = \frac{1}{4}(\parallel u + v\parallel^2 - \parallel u - v\parallel^2) - \frac{\imath}{4}(\parallel u + \imath v\parallel^2 - \parallel u - \imath v\parallel^2)

Using the definition of the inner product, we can rewrite this as:

<u|v> = \frac{1}{4}((u+v)*(u+v) - (u-v)*(u-v)) - \frac{\imath}{4}((u+\imath v)*(u+\imath v) - (u-\imath v)*(u-\imath v))

Expanding the brackets and simplifying, we get:

<u|v> = \frac{1}{4}(u*u + u*v + v*u + v*v - u*u + u*v - v*u + v*v) - \frac{\imath}{4}(u*u + u*\imath v + \imath u*v + \imath^2 v*v - u*u + u*\imath v - \imath u*v + \imath^2 v*v)

Since \imath^2 = -1, this simplifies to:

<u|v> = \frac{1}{4}(2u*v + 2v*u) - \frac{\imath}{4}(2u*\imath v + 2\imath u*v)

Using the properties of the inner product, we can write this as:

<u|v> = \frac{1}{4}(2(u*v + v*u)) - \frac{\imath}{4}(2\imath(u*v - v*u))

Simplifying further, we get:

<u|v> = \frac{1}{2}(u*v + v*u) - \frac{\imath}{2}(u*v - v*u)

Finally, using the