Question

Let u and upsilon be vectors in an inner product space. Show that ||u + upsilon||^2 + ||u - upsilon||^2 = 2 (||u||^2 + ||upsilon||^2).

Answer 1

Answer with Step-by-step explanation:

We are given that u and upsilon be vectors in an inner product space .

We have to show that $\left \| u+upsilon \right \|^2+\left \|u-upsilon \right \|^2=2(\left \| u\right \|^2+\left \| upsilon \right \|^2)$

We know that

$\left \| u+upsilon \right \|^2=<u+upsilon,u+upsilon>= \left \| u \right \|^2+ \left \| upsilon \right \|^2+2 \left \| u\right \| \left \| upsilon \right \|$

$\left \| u+upsilon \right \|^2=< u-upsilon,u-upsilon>=\left \| u \right \|^2+ \left \| upsilon \right \|^2-2 \left \|u \right \| \left \| upsilon \right \|$

Left hand side

Using above identities

$\left \| u+upsilon \right \|^2+\left \| u-upsilon \right \|^2$

$=<u+upsilo,u+upsilon >+< u-upsilon,u-upsilon>$

$=\left \| u \right \|^2+ \left \| upsilon \right \|^2+2 \left \|u \right \| \left \| upsilon \right \|+\left \| u \right \|^2+ \left \| upsilon \right \|^2-2 \left \|u \right \| \left \| upsilon \right \|$

$=2\left \| u \right \|^2+2\left \| upsilon \right \|^2$

$=2(\left \|u \right \|^2+\left \| upsilon \right \|^2$

Hence, proved.