Answer to Question #208118 in Linear Algebra for Faith

Question

1.find an expression for 1/2||u+v||²+1/2||u-v||² in terms of ||u||²+||v||².

2.find an expression for ||u+v||²-||u-v||² in terms of u×v.

3.use the result of 2 to deduce an expression for ||u+v||² whenever u and v are orthogonal to each other.

Answer 1

1. Given arbitrary vectors $u$ and $v.$ We get

$||u+v||^2=(u+v)(u+v)$

$=(u,u)+(u,v)+(v,u)+(v,v)$

$=||u||^2+2(u,v)+||v||^2$

$||u-v||^2=(u-v)(u-v)$

$=(u,u)-(u,v)-(v,u)+(v,v)$

$=||u||^2+2(u,v)+||v||^2$

Then

$||u+v||^2+||u-v||^2$

$=||u||^2+2(u,v)+||v||^2+||u||^2-2(u,v)+||v||^2$

$=2||u||^2+2||v||^2$

Therefore

$\dfrac{1}{2}||u+v||^2+\dfrac{1}{2}||u-v||^2=||u||^2+||v||^2$

2.

$||u+v||^2-||u-v||^2$

$=||u||^2+2(u,v)+||v||^2-||u||^2+2(u,v)-||v||^2$

$=4(u,v)$

Therefore

$||u+v||^2-||u-v||^2=4u\cdot v$

3. If  u and v are orthogonal to each other, then $u\cdot v=0.$

Hence

$||u+v||^2-||u-v||^2=0$

$=>||u+v||^2=|u-v||^2$

$\dfrac{1}{2}||u+v||^2+\dfrac{1}{2}||u-v||^2$

$=\dfrac{1}{2}||u+v||^2+\dfrac{1}{2}||u+v||^2$

$=||u+v||^2$

$=\dfrac{1}{2}||u-v||^2+\dfrac{1}{2}||u-v||^2$

$=||u-v||^2$

$=||u||^2+||v||^2$

$||u-v||^2=||u+v||^2=||u||^2+||v||^2$