2
$\begingroup$

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: $(1,0,1), (0,1,-1)$?

$\endgroup$
1
$\begingroup$
  1. 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.

  2. 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.

$\endgroup$
2
  • $\begingroup$ For 2. I came up with (1,1,0) and (0,1,1) as the basis for $\mathbb{R}^{3}$. Even though they're linearly independent, I can't write, say, vector (3,3,3) as a linear combination of (1,1,0) and (0,1,1), right? So this means they're not a basis? $\endgroup$
    – Adrian
    Feb 10 '14 at 2:41
  • $\begingroup$ Exactly; they don't span $\mathbb{R}^3$, so don't form a basis of $\mathbb{R}^3$. $\endgroup$ Feb 10 '14 at 3:58
1
$\begingroup$
  1. 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.

  2. Both of your examples are correct so far as I can tell.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.