W={(x,y,z)} in R3: |x|+|y|=|z|}
It seems as if though this would be a subspace but apparently it's not and I would like to know why.
It seems as if though this would be a subspace but apparently it's not and I would like to know why.
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Usually, subspaces are associated with linear equations without nonzero constant terms. The absolute value makes the equation nonlinear, which suggests that W might not be a subspace. So it makes sense to try to show that W is not a subspace.
Recall that a subspace must contain the zero vector, be closed under addition, and be closed under scalar multiplication. Violation of just one of these three conditions is enough for a set not to be a subspace.
Since the sum of the absolute values of two numbers is not always the absolute value of the sum, this suggests that we try to show that closure under addition is violated.
We can produce a counterexample to show that W is indeed not closed under addition.
Note that (1, 1, 2) and (1, -1, 2) are in W, since they both satisfy |x| + |y| = |z|.
However, their sum, (2, 0, 4) is not in W, since it fails to satisfy |x| + |y| = |z|.
So W indeed violates closure under addition and therefore is indeed not a subspace!
Lord bless you today!
Recall that a subspace must contain the zero vector, be closed under addition, and be closed under scalar multiplication. Violation of just one of these three conditions is enough for a set not to be a subspace.
Since the sum of the absolute values of two numbers is not always the absolute value of the sum, this suggests that we try to show that closure under addition is violated.
We can produce a counterexample to show that W is indeed not closed under addition.
Note that (1, 1, 2) and (1, -1, 2) are in W, since they both satisfy |x| + |y| = |z|.
However, their sum, (2, 0, 4) is not in W, since it fails to satisfy |x| + |y| = |z|.
So W indeed violates closure under addition and therefore is indeed not a subspace!
Lord bless you today!