0
$\begingroup$

I seem to always have troubles when starting proofs. My professor said that the proofs he gave us today are mostly one line proofs, but I just don't know where to start with this one.

What I've tried: $\langle u,v\rangle=(x_1,x_2,\ldots x_n)\cdot(y_1,y_2,\ldots y_n)=x_1y_1+x_2y_2+\cdots +x_ny_n$

and then I don't know where to go from there or if I'm even on the right track.

$\endgroup$
2
$\begingroup$

You can show this as follows;

expanding $||x+y||^2=\langle x+y,x+y \rangle=||x||^2+2\langle x, y \rangle + || y||^2$

if you want to say why this is, then it is just from the properties of inner product. i.e., $$\langle x+y , x+y \rangle= \langle x, x+y \rangle + \langle y, x+y \rangle$$ (linearity)

$$= \langle x+y , x \rangle + \langle x+y , y \rangle$$ (symmetric property)

$$= \langle x , x \rangle + \langle y , x \rangle + \langle x , y \rangle + \langle y ,y \rangle$$

(Linearity again)

and similarly

$|| x-y || ^2= \langle x-y,x-y \rangle= ||x|| ^2 - 2 \langle x, y \rangle + || y||^2$

Now just subtract these two equations to obtain the final result.

ie

$$|| x +y ||^2-||x-y||^2=4\langle x , y \rangle$$

ie, $$\langle x , y \rangle=\frac{1}{4}(|| x + y||^2-||x-y||^2)$$

You can make it one line or so by just expanding and saying subtract etc, if you are not required to justify each and every step. Of course the result holds if x and y were vectors in $\mathbb{R^n}$ . and this is also known as the "polar form" by some.

$\endgroup$
  • $\begingroup$ Thanks a lot for the thorough explanation of how to solve this problem. Kudos sir! $\endgroup$ – Linear Algebra Noob Apr 29 '15 at 8:09
1
$\begingroup$

Note that in a real vector space we have that: $$ <u,v> = <v,u>$$ $$<u+v,w> = <u,w> + <v,w>$$ $$ <\alpha u,v > = \alpha <u,v>, \mbox{ for } \alpha \in \mathbb R.$$

Using these facts we see that: $$ \lVert u+v \lVert^2<u+v,u+v> = 2<u,v>+<u,u>+<v,v> $$ $$ \lVert u-v \lVert^2 = -2<u,v>+<u,u>+<v,v>$$

Inserting this on the right side of your formula we get the desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.