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## Homework Statement

{(x

_{1},x

_{2})

^{T}| |x

_{2}|=|x

_{2}|}

So my first thought is we would have to check for both cases (x1,x1) and (-x1,-x1)

a=(x1,x1)

^{T}b=(v1,v1)

^{T}

βa=(βx1,βx1)

^{T}for the case where a<0 βa(-βx1,-βx1)

^{T}

thus it is closed under scalar multiplication.

a+b=(x1+v1,x1+v1) for the case where a and b<0 [-(x1+v1,x1+v1)]

^{T}

thus closed under addition.

I would think this would be a subset, however the book says this is false. Where did I go wrong?