If $\{x,y\}$ is an orthonormal set in an inner product space then the value of $\|x-y\|+\|x+y\|$ is $2\sqrt2$.

Question

If $\{x,y\}$ is an orthonormal set in an inner product space then the value of $\|x-y\|+\|x+y\|$ is $2\sqrt2$.

Using $\langle x,y \rangle=0$ and $\langle x,x \rangle=\langle y,y \rangle=1$ and the definition of inner product I can expand this expression and get the answer but I was wondering if there's a faster method since I'm only given a few minutes for this MCQ.

Answer 1

Hint By Pytagoras Theorem in an inner product space $$\|x \pm y\|^2=\|x\|^2+\|y\|^2=1+1=2$$

P.S. You can also figure the answer by "drawing a diagram" of two orthonormal vectors... But one needs to be carefull, this approach works here but may lead to the wrong answer in some situations, it is a heuristic and not formal approach.

Answer 2

Your numerical calculation is essentially correct, and about as fast as possible.

I think you mean $x = \langle 1, 0 \rangle$ and $y = \langle 0, 1\rangle$. Then each of $x \pm y$ is a vector of length $\sqrt{2}$.

For a one line more theoretical argument see the answer from @N.S.