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

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2 Answers 2

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  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.

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  • $\begingroup$ 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
    Commented Feb 10, 2014 at 2:41
  • $\begingroup$ Exactly; they don't span $\mathbb{R}^3$, so don't form a basis of $\mathbb{R}^3$. $\endgroup$ Commented Feb 10, 2014 at 3:58
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  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.

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