Elaboration of an example on the independence of the conditions of a subspace of a vector space.

Question

The example that is given in Golan's "Linear Algebra" is given below:

My questions are:

1- Why the first set is not closed under scalar multiplication?

2-Why the second set is closed under scalar multiplication? and why it is not closed under vector addition?

Thanks!

Answer 1

1. Consider the vector $x = (1,0,0)$ which is contained in the first set. If you scale it by $\frac 1 2$, you get $\frac 1 2 x = (\frac 1 2,0, 0)$, which is not contained in the set.
2. If $abc = 0$, then at least one of the three numbers $a, b$ or $c$ has to be zero. If you scale all of them, then this does not change, because $\lambda \cdot 0 = 0$. To see that it is not closed under addition, consider the vectors $x = (1,0,0)$ and $y = (0,1,1)$, both are contained in the set. But $x + y = (1,1,1)$, which does not have the desired property.