Inner product space

**StefanM** · March 12th 2011, 05:10 AM

How ca I prove that in a real/complex vector space V
$||x + y||^{2}+ ||x - y||^{2} = 2(||x||^{2} + ||y||^2)$
Also If I consider the parallelogram
with adjacent sides OP; OQ where P is the point (x1; x2); Q is the point (y1; y2) and O is
the origin.What does this say about a parallelogram in the plane?
This is an exercisse from LINEAR ALGEBRA AND ITS APPLICATIONS David C. Lay.

**Plato** · March 12th 2011, 05:24 AM

Originally Posted by StefanM

How ca I prove that in a real/complex vector space V
$||x + y||^{2}+ ||x - y||^{2} = 2(||x||^{2} + ||y||^2)$

Just consider this: $\|x+y\|^2=(x+y)\cdot(x+y)$ .
Now expand to get $x\cdot x+2x\cdot y+y\cdot y.$
Do that for $\|x-y\|^2$ and add.

**StefanM** · March 12th 2011, 05:44 AM

I was confuse about the fact that $||x^{2}||=<x,x>$ ...How can I solve the parallelogram problem?

**Plato** · March 12th 2011, 06:04 AM

Originally Posted by StefanM

I was confuse about the fact that $||x^{2}||=<x,x>$ ...How can I solve the parallelogram problem?

You asked about proving $\|x+y\|^2+\|x-y\|^2=\|x\|^2+2\|x\|\|y\|+\|y\|^2$ .
That what I was addressing.
It is a fact that $\|x+y\|^2=<x+y,x+y>$ , which can be expanded.

$\|x+y\|~\&~\|x-y\|$ are the lengths of the diagonals of a parallelogram in the plane.

Thread: Inner product space

LinkBack

Thread Tools

Display

Inner product space

Similar Math Help Forum Discussions

[SOLVED] Need clarification:Definition of Euclidean n-space and Inner Product Space

Every vector space over R or C can be made into an inner product space

Inner Product Space

closed in dual space E* implies closed in product space F^E

Inner Product Space

Search Tags