$||\underline{x}+\underline{y}||= ||\underline{x}-\underline{y}||$ implies x is perpendicular to y.

Question

Let $(X,\langle\cdot,\cdot\rangle)$ be an inner product space. Prove if $X$ is a real vector space, then $\lVert x+y\rVert= \lVert x-y \rVert$ implies $x\perp y$.

Answer 1

Hint: we can write $$ \|x+y\|^2 = \|x-y\|^2 \implies \langle x+y,x+y\rangle = \langle x-y,x-y\rangle $$ Using the linearity of an inner-product, expand the above.

Answer 2

By the definition of norms $$\|x+y\|^2 = \|x-y\|^2 \implies \langle x+y,x+y\rangle = \langle x-y,x-y\rangle. $$

Simplifying, we'll get $$\langle x,y\rangle=0$$

Answer 3

Since $\|x+y\|^2=\langle x+y,x+y \rangle$ and $\|x-y\|^2=\langle x-y,x-y \rangle$ then if they are equal you obtain $$\langle x,x\rangle +\langle y,y \rangle + \langle x,y \rangle + \langle y,x\rangle = \langle x,x\rangle +\langle y,y \rangle - \langle x,y \rangle - \langle y,x\rangle $$ and so $$2\langle x,y\rangle = -2\langle y,x\rangle .$$

Answer 4

The parallelogram spanned by $x$ and $y$ has equal diagonals, hence it is a rectangle. “Elementary, Watson.”