Finding a set of vectors that is not a vector space I am asked to find a set of vectors in $\mathbb{R}^2$ such that if $y$ is in the set, then $b\cdot y$ is in the set for every real number $b$. However, I am told that this set cannot be a vector space.
The set would not be a vector space if the zero vector weren't included. However, given that $b$ can be any real number, including $0$, this cannot be the case. How else can a set of vectors that matches this condition not be a vector space?

Let $S$ be the union of the first and third quadrants (along with the $x$- and $y$-axes, let's say).  Scaling by a real number $r$ will either keep each vector in its same quadrant (if $r>0$), or send the first and third quadrants to each other (if $r<0$), or send all vectors to the origin (if $r=0$).  So $S$ is closed under scalar multiplication.
However, notice that
$$
(-2, -1) + (1, 2) = (-1, 1)
$$
which is in the second quadrant, so $S$ is not closed under vector addition.
Extra note: for an example of something closed under addition but not under scalar multiplication, take just the first quadrant.
