# Inner product space parallelogram law? Kind of confused what is being asked here. Isn't this just obviously true? How would I start this proof?

• It's true for norms defined with an inner product, which is the case for the euclidean norm (and which is why you find it obvious). It's not necessarily true for other norms. – Bernard Apr 28 '15 at 23:03

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 add these two equations to obtain the final result.

ie, $$||x+y||^2+||x-y||^2=2||x||^2+2||y||^2$$

In terms of a geometric interoperation in $\mathbb{R^2}$ think of two of the sides being the vector x and the other two sides being the vector y. The vectors $x+y$ and $x-y$ correspond to the diagonals. Photo from "Schaums linear algebra, 4th edition"

• @user139985 , is this not what you were looking for? – Quality Apr 28 '15 at 23:47

When proving the parallelogram law, yes prove it by expanding the norms and cancelling.

For the diagram, think about the corner of the parallelogram a the origin. Define the two neighbouring vertices as $a$ and $b,$ and the fourth as $a+b.$

The left and right sides of the parallelogram have length $|b|$ and the top and bottom sides of the parallelogram have length $|a|.$

If we express the diagonals in terms of $a$ and $b.$ Doing ths, the sum of the square of the lengths of the diagonals is $|a+b|^2+|b-a|^2$ and the sum of the square of the lengths of the sides is $2|a|^2+2|b|^2,$ and the law comes from the fact tat we are showing that these are equal.