Question

Linear algebra subspace problem?
How would you prove that this set does not form a subspace in R^2?
[(x1,x2)^T ; abs(x1) = abs(x2)}
I let x1 and x2 be some constant {c1, c1}.
When I multiple (c1 and c2) with any scalar by y.... y{c1,c1}={yc1,yc1} which is closed under multiplication.
And similar when I add {c1, c1} with {y1, y1}...it'll end up like {c1+y1} and (c1 + y1} which is closed under addition.
So how can it not be a subspace? Thanks for any replies

Answer 1

Consider the vectors x=(-1,1) and y=(1,1). Both these vectors satisfy abs(x1)=abs(x2) but adding them together gives (0,2) which does not satisfy the condition. Therefore this counterexample shows that the set is not closed under addition and so it is not a subspace.

Answer 2

hmm how come x can have negative values? I thought it could only have positive values since the condition states it wants the absolute value

Answer 3

The condition only states that the absolute value of x1 must equal the absolute value of x2. It doesn't say that x1 or x2 must themselves be positive.