why is $ A=\{(0,x,z)|x,z\in R\}$ is a two dimensional subspace space of $R^{3}$ over $R$ but $B=\{(0,0,z)|z\in R\} $ $\cup$ $\{(0,x,0)|x\in R\}$ is not?
i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and
B is direct sum of two subspace of $ R^{3}$ with same basis .
On
$B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)\notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $\mathbb R^3$.
To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $\mathbb R^2$, which you know to be a vector space.
$B$ is defined as the union of two subsets of $\mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $\mathbb R^{3}$.