Question

example of parallelogram law in inner product space

Answer 1

Answer:

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Step-by-step explanation:

example of parallelogram law in inner product space

The Parallelogram Identity for Inner Product Spaces

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The Parallelogram Identity for Inner Product Spaces

We will now look at an important theorem. If V is an inner product space and u,v∈V then the sum of squares of the norms of the vectors u+v and u−v equals twice the sum of the squares of the norms of the vectors u and v, that is:

(1)

(u+v)2+(u−v)2=2(u)2+2)v)2

This important identity is known as the Parallelogram Identity, and has a nice geometric interpretation is we're working on the vector space R2:

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We will now prove this theorem.

Theorem 1 (The Parallelogram Identity): Let V be an inner product space.

If u,v∈V then ∥u+v∥2+∥u−v∥2=2∥u∥2+2∥v∥2.

Proof: Let V be an inner product space and let u,v∈V. Noting that −1¯¯¯¯=−1 and we have that:

(2)

(u+v)2+(u−v)2=<u+v,

u+v>+<u−v,u−v>(u+v)2+(u−v)2=<u,

u>+<u,

v>+<v,

u>+<v,

v>+<u,

u>+<u,

−v>+<−v,

u>+<−v,

−v>(u+v)2+(u−v)2=2<u,

u>+2(−1)(−1¯¯¯¯<−v,

−v>+<u,

v>+<v,

u>−1¯¯¯¯<u,v>−<v,

u>(u+v)2+)u−v)2=2<u,

u>+2<v,

v>+<u,

v>+<v,

u>−<u,

v>−<v,

u(u+v)2+(u−v)2=2(u)2+2(v)2■

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