# Orthogonality

• Nov 28th 2007, 06:41 PM
pakman
Orthogonality
If x=(x1,x2)^T, y=(y1,y2)^T, and z=(z1,z2)^T are arbitrary vectors in R^2 prove that

a) x^Tx (x transpose x) >= 0
So basically I got x1^2 + x2^2 >= 0

Not sure where to go with this proof... thanks
• Nov 28th 2007, 07:15 PM
ThePerfectHacker
Quote:

Originally Posted by pakman
If x=(x1,x2)^T, y=(y1,y2)^T, and z=(z1,z2)^T are arbitrary vectors in R^2 prove that

a) x^Tx (x transpose x) >= 0
So basically I got x1^2 + x2^2 >= 0

Not sure where to go with this proof... thanks

Because $x_1^2\geq 0\mbox{ and }x_2^2\geq 0$ because they are squares.