Is $B=(1,1,0) (1,0,1) (0,1,1)$ a basis of $R^3$? Is $B=\{(1,1,0) (1,0,1) (0,1,1)\}$ a basis of $\mathbb R^3$?
I am doubting how this vector would span $\mathbb R^3$, for example: how will the vector $(2,0,0)$ be in the span of $B$? 
I am a beginner, just finished studying bases. This doubt might be very basic, I tried to find on internet but I could not understand much of it. I was looking for some lucid explanation. Thanks.
 A: When you have $n$ vectors in $\mathbb{R}^n$, they span the space if and only if they are linearly independent. When you have fewer than $n$ vectors in $\mathbb{R}^n$ they'll never span the space, and when you have more than $n$ vectors in $\mathbb{R}^n$ they'll span the space if there is a subset of size $n$ that is linearly independent. A basis is a spanning set that is also linearly independent, which can only happen when the number of vectors equals the dimension of the space.
You can use any technique you like to see that these vectors are indeed linearly independent (row reduction is popular). Since they are linearly independent and the dimension of the vectors space equals the number of vectors, they form a basis.
For the specific question about $(2,0,0)$, we see that $$(2,0,0)=(1,1,0)+(1,0,1)-(0,1,1)$$ This can be found by solving the equation $$(2,0,0)=a(1,1,0)+b(1,0,1)+c(0,1,1)$$ Although this looks like one equation with three unknowns, we can break it into three equations by looking at each coordinate separately, such as $2=a\cdot 1+b\cdot 1+c\cdot 0$. Once you've broken this into a system of equations, you can solve it via your favorite technique.
A: Yes, because for arbitrary $(a_1,a_2,a_3)\in\mathbb{R}^3$ we have $(a_1,a_2,a_3)=\frac{a_1+a_2-a_3}{2}(1,1,0)+\frac{a_1+a_3-a_2}{2}(1,0,1)+\frac{a_2+a_3-a_1}{2}(0,1,1)$. The coefficients can be achieved by setting $(a_1,a_2,a_3)=x(1,1,0)+y(1,0,1)+z(0,1,1)$ and solving $(x,y,z)$ in terms of $(a_1, a_2, a_3)$.
A: $B=\{(1,1,0),(1,0,1),(0,1,1)\}$ is a base for $\Bbb R^3$. 
To verify that, $\forall (x,y,z)\in \Bbb R^3$ it must be true that $\exists\{\alpha_1,\alpha_2,\alpha_3\}\subset \Bbb R$ such that
   $$(x,y,z)=\alpha_1 (1,1,0)+\alpha_2 (1,0,1)+\alpha_3 (0,1,1)$$
By solving a system of equations you will find that
$$\alpha_1=\frac{x+y-z}{2},\ \ \alpha_2=\frac{x-y+z}{2},\ \  \alpha_3=\frac{y-x+z}{2}$$
will do the trick. For instance, when $(x,y,z)=(2,0,0)$, use $\alpha_1=1$, $\alpha_2=1$ and  $\alpha_3=-1$. 
It is not possible to show that a set of vectors, such as B, is a base for $\Bbb R^3$ just checking if they span a particular instance from $\Bbb R^3$, such as $(2,0,0)$, this set must span all vectors in $\Bbb R^3$.
