Let $x$ and $y$ be unit vectors in $\mathbb{R}^n$, Such that $x+y$ and $x-y$ are non-zero. Show that $x+y$ and $x-y$ are perpendicular.

Question

I try it graphically but I don't think its the correct one.

Answer 1

Let $x=(x_1,x_2,..,x_n)$ and $y=(y_1,y_2,..,y_n)$ then, $\sum_{k=1}^{k=n}{x_k^2}=1$, $\;$ $\sum_{k=1}^{k=n}{y_k^2}=1$, $\;$ $x-y=(x_1-y_1,x_2-y_2,..,x_n-y_n)$ and $x+y=(x_1+y_1,x_2+y_2,..,x_n+y_n)$. With dot product, $0=\sum_{k=1}^{k=n}{(x_k^2-y_k^2)}=(x-y).(x+y)=|x||y|\cos\alpha=\cos\alpha$ (here $\alpha$ is the angle between $x$ and $y$) thus, $\alpha=\pi/2$

Answer 2

Using properties of the dot product, $(x+y)\cdot(x-y)=x\cdot x - x\cdot y + y \cdot x - y \cdot y=x\cdot x - y\cdot y. $

Furthermore, if $x$ and $y$ are unit vectors, then $x\cdot x=y\cdot y=1,$ so $(x+y)\cdot(x-y)=0. $

What does that mean?