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