Understanding of Example of Supspace Let $V$=$K^{n}$ and let $W$ be the set of vectors in $V$ whose last coordinate is equal to $0$. Then $W$ is a subspace of $V$, which we could identify with $K^{n-1}$.
My question is that why last coordinate is equal to $0$? Can it be every time is equal to $0$? Finally, Why we could identify with $K^{n-1}$ (I didn't understand really)?
 A: We are defining $W$ to be all of the vectors in $V$ with $0$ last coordinate. You can check that if you multiply a vector $w \in W$, then it still has last coordinate $)$, and if you add two vectors $w, w' \in W$, then the sum$w + w'$ still has last coordinate $0$. Hence $W$ is a subspace.
When we say that we can identify $W$ with $K^{n - 1}$, we mean there is a linear transformation $T : K^{n - 1} \to W$ that is both injective and surjective. Such a transformation is called an isomorphism. It means that $W$ and $K^{n - 1}$, although the elements in them are not precisely equal, the two vector spaces have the same exact structure, and are thus essentially the same.
A: It is always best to consider examples. Let $V = \mathbb{R}^3$ and $W = \{(x,y,z): z = 0\}$ then we can identify $W$ with $\mathbb{R}^2 = \mathbb{R}^{3-1}$. 
So in general if $V = \mathbb{K}^n = \{(k_1,...,k_n): k_i \in K\}$ and $W = \{(k_1',...,k_n'): k_i \in K, k_n' = 0\}$ then there is an isomorphism between $W$ and $\mathbb{K}^{n-1}$ namely,
$$(k_i',...,k_{n-1}',0) \mapsto (k_i',....,k_{n-1}')$$
