Why a basis of the vector space $V=\{0\}$ is $B=\{\}$? I know that every $\vec{v}\in V$ can be written as $\vec{v}=a_1\vec{u_1}+...+a_n\vec{u_n}$, where $B=\{\vec{u_1},...,\vec{u_n}\}$ is a basis of the vector space. But I can't understand why a basis of the vector space $V=\{0\}$ is $B=\{\}$.
 A: Your question can be rephrased as: 


*

*Why $\text{span}(\varnothing)=\{0\}$?

*Why $\varnothing$ is linearly independent?
Note that span of a set is the intersection of all subspaces containing it. So here intersection of all subspaces containing $\varnothing$ is $\{0\}$.
For the second one, suppose empty set is dependent. Then there are scalars so that the linear combination of elements of $\varnothing$ is zero. But in $\varnothing$, there is no element at all, so this cannot be happen and hence empty set is independent
Hence empty set is a basis for the trivial space
A: To say that $B\subseteq V$ is a basis for a vector space $V$ is to say that (1) $B$ spans $V$, and (2) $B$ is linearly independent. 
Unpacking the definitions of (1) and (2), we notice that for both, we have to think about linear combinations of elements of $B$. In general, if $B = \{b_1,\dots,b_n\}$, then a linear combination of elements from $B$ is something of the form $$\sum_{i=1}^n c_ib_i$$ where the $c_i$ are scalar coefficients. Note that there is one term in the sum for each element of $B$. 
Ok, what happens when $B$ is empty? Then there are zero terms in the sum. And it is an important principle that the sum of zero terms is the additive identity $0$ (likewise, the product of zero terms is the multiplicative identity $1$). So there is exactly one linear combination of elements of $B$, and it is the vector $0$. 
Ok, why is (1) true? Well, to say $B = \varnothing$ spans $V = \{0\}$ is to say that every vector $v\in V$ is a linear combination of elements from $B$. But the only vector in $V$ is $v = 0$, and we've just seen that $0$ is the empty linear combination of elements of $B$. 
And why is (2) true? To say that $B$ is linearly independent means that the only way we can write $0$ as a linear combination of elements of $B$ $$\sum_{i=1}^n c_ib_i = 0,$$ is if all of the scalar coefficients $c_i$ are $0$. We noted above that there is exactly one way to write $0$ as a linear combination of elements of $B$ (as the empty sum). Is it true that in the empty sum, all of the scalar coefficients $c_i$ are $0$? Yes, because there aren't any scalar coefficients. 
(On the last point: it's another important principle that when there are no things of a certain kind, then all things of that kind satisfy any property you like. For example, all negative real numbers which have a real square root are integers. Also, all negative real numbers which have a real square root are unicorns. This is called vacuous truth.)
A: $\{0\}$ is $0$-dimensional:  there are no elements in a basis. 
