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

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

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

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

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

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

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

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

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

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