The clue here that this is not a subspace is the presence of the 2 in the second component. Typically for a subspace, we expect to see linear expressions involving our variables, but in linear algebra, the adjective
linear doesn’t imply the inclusion of constant terms the way it does in calculus. The reason, again, is the special role of zero in a vector space.
While it’s true that this set doesn’t contain the zero vector (which rules it out as a subspace), it’s not as obvious: perhaps there are values of
and
that give us
and
as well? Solving a system of equations would tell us that indeed, this is not possible.
We could also show that the closure conditions fail. Putting
gives the element
and putting
gives the element
Adding these, we get the vector
Why is this not in the set? We would need
so
Then
implies
but