What are the basis that span $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ I have two questions: 1. What are the basis that span $\mathbb{R}^{2}$ Is it just $(0,1)$ and $(1,0)$? I read somewhere that it is $(0,1), (1,0), (1,1)$. But the 3rd one can be written as a linear combination of the first two, so would it still be considered as a basis? 
2.Find a basis for a two-dimensional subspace of $\mathbb{R}^{3}$ that does not contain $(1,0,0), (0,1,0), (0,0,1)$. I came up with $(1,1,0)$ and $(0,1,1)$ Is this correct? I can think of a couple others too: $(1,0,1), (0,1,-1)$?
 A: *

*It's true that the second set given in your first question spans $\mathbb{R}^2$, but to be a basis a set must be a minimal spanning set; i.e., there cannot be an element in it whose removal renders the set still a spanning set. Since, as you rightly pointed out, the third vector in that set can be removed and the result still spans, you know that it's not a basis.

*Both of your examples are correct so far as I can tell.
A: *

*The set $$\{(0,1),(1,0)\}$$ is a span of $\mathbb{R}^2$.  It has the two key properties:


*

*it spans $\mathbb{R}^2$, i.e., any vector in $\mathbb{R}^2$ can be formed by a linear combination of the vectors in the set, and

*it is linearly independent.


The set $$\{(0,1),(1,0),(1,1)\}$$ is not a basis for $\mathbb{R}^2$.  It spans $\mathbb{R}^2$, but it is not linearly independent, which can be demonstrated by identifying a linear dependency, e.g. $$(0,1)+(1,0)-(1,1)=(0,0).$$
However, there are other bases for $\mathbb{R}^2$ aside from the one above.  E.g. $$\{(0,1),(1,1)\}$$ is another basis.

*In order to show they are correct, we need to check they are linearly independent.  For two vectors, it's sufficient to "inspect" that one is not a scalar multiple of the other.
